## LaTeX forum ⇒ Text Formatting ⇒ Problem aligning

Information and discussion about LaTeX's general text formatting features (e.g. bold, italic, enumerations, ...)
danielvelizv
Posts: 45
Joined: Wed May 25, 2016 7:04 am

### Problem aligning

Hello Stefan, from the document I posted before, the code:

\documentclass[10pt,letterpaper,fleqn]{scrbook}\usepackage[left=3cm,right=2.5cm,top=2.5cm,bottom=2.5cm,  marginparwidth=2.85cm, marginparsep=0pt]{geometry}\usepackage[leqno]{mathtools}\newtagform{bold}{}{\textbf{)}}\usetagform{bold}\renewcommand*{\theequation}{\textbf{Ej. 1.}}\newcommand{\remark}[1]{& \quad \longleftarrow  & \quad\fontsize{8}{0}\selectfont\parbox{0.25\textwidth}{\raggedright#1}}\setlength{\jot}{10pt}\begin{document}\begin{equation}\begin{alignedat}[t]{2}  \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx      &= \int \frac{x^2}{\sqrt{x}}\,dx         + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx         \remark{separar en 3 integrales} \\      &= \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx         + \int \frac{1}{x^{1/2}}\,dx         \remark{escribir los radicales en forma de potencia} \\      &= \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx         \remark{reescribir} \\      &= \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C         \remark{integrar} \\      &= \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} +         \remark{simplificar}    \end{alignedat}\end{equation}\end{document}

produces the following result in the attached picture, the bf text must to be left-aligned in the first line, I don't know what's wrong here...this is the recommendation you gave me to align the equations!
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Stefan Kottwitz
Posts: 9550
Joined: Mon Mar 10, 2008 9:44 pm
Did you test this code? Because that code positions the bold number at the left. Maybe you tested it within your main document with different settings.

Here, the point is the leqno option. leqno means left equation numbering. I added it to mathtools:

\usepackage[leqno]{mathtools}

but it also can be added to amsmath

\usepackage[leqno]{amsmath}

or to the documentclass as option. You load both packages. I suggest: add leqno as option in the first line of your document, to the class. The packages will inherit it.

Stefan

danielvelizv
Posts: 45
Joined: Wed May 25, 2016 7:04 am
Stefan_K wrote:Did you test this code? Because that code positions the bold number at the left. Maybe you tested it within your main document with different settings.

Here, the point is the leqno option. leqno means left equation numbering. I added it to mathtools:

\usepackage[leqno]{mathtools}

but it also can be added to amsmath

\usepackage[leqno]{amsmath}

or to the documentclass as option. You load both packages. I suggest: add leqno as option in the first line of your document, to the class. The packages will inherit it.

Stefan

Yes, I did it, I tested your code and followed your instructions but now, my code:

     \documentclass[10pt,letterpaper,leqno,     headsepline,footsepline,     plainheadsepline,plainfootsepline,     ]{scrbook}     \usepackage[left=3cm,right=2.5cm,top=2.5cm,bottom=2.5cm, marginparwidth=2.85cm, marginparsep=0pt,head=22.22223pt]{geometry}     \usepackage[utf8]{inputenc}     \usepackage[spanish]{babel}     \usepackage{amsmath}     \usepackage{mathtools}            %MANIPULACIÓN DE LA ALINEACIÓN LATERAL DE LAS EXPRESIONES MATEMÁTICAS     \usepackage{amsfonts}     \usepackage{amssymb}     \usepackage{graphicx}     \usepackage[most]{tcolorbox}     \usepackage{xcolor}     \usepackage{tikz}     \usepackage{array}     \usepackage{marginnote}           %COLOCACIÓN DE NOTAS DE PÁGINA EN LOS LADOS     \usepackage{setspace}             %SEPARACIÓN DE LÍNEAS EN PÁRRAFOS     \usepackage{cancel}               %CANCELACIÓN DE TÉRMINOS     \usetikzlibrary{calc}     \usetikzlibrary{shapes.callouts}  %CUADROS DE IDEAS     \usetikzlibrary{decorations.text}     \usetikzlibrary{positioning}     \usepackage{varwidth}     \def\cabecera#1{       \begin{tikzpicture}[overlay, remember picture]         \draw let \p1 = (current page.west), \p2 = (current page.east) in           node[minimum width=\x2-\x1, minimum height=3cm, line width=0pt, rectangle, fill=gris!80, anchor=north west, align=left, text width=17cm] at ($(current page.north west)$)    {#1};       \end{tikzpicture}     } 	\newtagform{bold}{}{\textbf{)}}	\usetagform{bold}	\renewcommand*{\theequation}{\textbf{Ej. 1.}}	\newcommand{\remark}[1]{& \quad \longleftarrow   & \quad\fontsize{8}{0}\selectfont\parbox{0.25\textwidth}{\raggedright#1}}	\setlength{\jot}{10pt}      \newcommand{\titulo}{{650 integrales indefinidas \\ resueltas ¡paso a paso!}}     \newcommand{\inmediata}{{Integrales inmediatas}}     \usepackage{scrlayer-scrpage}     \addtokomafont{pagenumber}{\bfseries}     \addtokomafont{pageheadfoot}{\fontsize{8}{9}\sffamily\upshape}     \clearpairofpagestyles     \ohead*{\inmediata}     \ihead*{\titulo}     \ofoot*{\pagemark}     \ifoot*{Ing. Daniel A. Veliz V.}        \setlength{\parindent}{0pt}      %SIN SANGRÍA EN LOS PÁRRAFOS     \setlength{\arraycolsep}{4pt}    %ANCHO DE LAS COLUMNAS EN LOS ARRAY     \setlength{\tabcolsep}{4pt}      %ANCHO DE LAS COLUMNAS EN LAS TABLAS     \setlength{\mathindent}{1pt}     %SIN SANGRÍA EN LA ALINEACIÓN MATEMÁTICA     \usepackage{anyfontsize}       %---------------------------------------------     %              COLORES DEFINIDOS     %---------------------------------------------       \definecolor{naranja}{rgb}{1, 0.3, 0}     \definecolor{blanco}{rgb}{0.97, 0.97, 1}     \definecolor{gris}{rgb}{0.47, 0.53, 0.6}     \definecolor{azul}{rgb}{0.12, 0.56, 1.0}     \definecolor{verde}{rgb}{0.0, 0.65, 0.31}     \definecolor{carmin}{rgb}{1.0, 0.0, 0.22}       \begin{document}     \cabecera{\bfseries {\fontsize{20}{0}\selectfont  \hfill Capítulo I \\ \hfill Integrales inmediatas}}     \begin{minipage}[c]{1\textwidth}     \vspace*{1cm}     En este capítulo se darán a conocer los fundamentos básicos de la integración de distintas funciones por medio del empleo de las propiedades matemáticas y así convertir las funciones integrando dadas en algunas de las formas básicas presentadas antes del desarrollo de este capítulo, de esta manera a medida que revise los capitulos posteriores se dará cuenta que la idea básica de aplicar las técnicas de integración consistirá en convertir integrandos complicados en formas elementales para determinar una \emph{función primitiva} o \emph{antiderivada} de una función $f$.     \\[0.5cm]     La antiderivación (o integración indefinida) se denota mediante el signo integral     $\displaystyle \int$ por lo tanto, el siguiente esquema podrá ayudarlo a identificar los elementos implícitos en el cálculo integral y qué se obtiene al calcular una integral indefinida:     \end{minipage}     \\[0.8cm]     \hspace*{3.75cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=azul!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0.7,-1)}, callout pointer width=5mm] {\begin{varwidth}{2cm} Función integrando \end{varwidth}};     \end{tikzpicture}     \hspace*{1.5cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=verde!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.5,-1)}, callout pointer width=5mm] {\begin{varwidth}{2.75cm} Antiderivada de la función $f$ \end{varwidth}};     \end{tikzpicture}     \\[-0.675cm]     \begin{equation}     \hspace*{5cm} \scalebox{1.5}{$\displaystyle \int \! \textcolor{azul!100}{f(x)}\,\textcolor{naranja!90}{dx} = \textcolor{verde!100}{F(x)} \ \textcolor{carmin!100}{+ \ C}$} \nonumber     \end{equation}     \\[-0.6cm]     \hspace*{5.85cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=naranja!90, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0,1)}, callout pointer width=5mm] {\begin{varwidth}{2cm}     Variable de integración \end{varwidth}};     \end{tikzpicture}     \hspace*{1.75cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=carmin!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.75,1)}, callout pointer width=5mm] {\begin{varwidth}{2.5cm}     Constante de integración \end{varwidth}};     \end{tikzpicture}     \\[0.75cm]     Además, según Larson R. (2009) en su texto \emph{Cálculo Integral - Matemáticas 2} expresa que:     \vspace{1ex}     \begin{quote}     La expresión $\displaystyle \int f(x)\,dx$ se lee como la antiderivada o primitiva de $f$ con respecto a $x$, el diferencial de $x$ sirve para identificar a $x$ como la variable de integración. El término \emph{integral indefinida} es sinónimo de antiderivada."     \end{quote}     \vspace{0.5cm}     \tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]     {\hspace*{-0.55cm} {\Huge {\bfseries\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{A lo largo de este texto, usted encontrará conforme vea las distintas técnicas y casos de integrandos particulares, la complejidad en el desarrollo de los mismos, como integrar cada función producirá una constante $C$, solo se asumirá en el resultado final escrito como la suma de todas las constantes de las integrales resueltas, de manera que $C = C_1 + C_2 + C_3 + \ldots + C_n$}     \vspace{0.6cm}     A continuación se presentará una lista de ejercicios con un orden aleatorio de dificultad y algunos ejemplos previamente explicados para ayudar a comprender el principio básico de la integración inmediata por medio del uso de la tabla.     \\[0.6cm]\begin{equation}\begin{alignedat}[t]{2}   \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx       &= \int \frac{x^2}{\sqrt{x}}\,dx          + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx          \remark{separar en 3 integrales} \\       &= \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx          + \int \frac{1}{x^{1/2}}\,dx          \remark{escribir los radicales en forma de potencia} \\       &= \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx          \remark{reescribir} \\       &= \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C          \remark{integrar} \\       &= \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} + C          \remark{simplificar}     \end{alignedat}\end{equation}           %\begin{tabular}{llllp{8cm}}     %\textbf{Ej. 1.1)} $\displaystyle \int x + 3\,dx$ & = & $\displaystyle \int x\,dx + \int 3\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[5mm]%                                   & = & $\displaystyle \int x\,dx + 3 \int dx$ & $\longleftarrow$ & %{\fontsize{8}{0}\selectfont en la integral de la derecha se extrajo el factor 3 fuera de la integral como una constante}     \\[1mm]%                   & = & $\displaystyle \frac{x^2}{2} + 3x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}  %   \end{tabular}     \newpage     \begin{tabular}{llllp{3cm}}     \textbf{Ej. 1.2)} $\displaystyle \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx$ & = & $\displaystyle \int \frac{x^2}{\sqrt{x}}\,dx + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]                                    & = & $\displaystyle \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx + \int \frac{1}{x^{1/2}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont escribir los radicales en forma de potencia}    \\[1mm]                                     & = & $\displaystyle \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]                                    & = & $\displaystyle \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar} \\[6mm]                                    & = & $\displaystyle \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}     \end{tabular}     \reversemarginpar     \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Las funciones irracionales (raíces) cuentan como funciones de potencia.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}     \\[0.9cm]     \begin{tabular}{llllp{4cm}}     \textbf{Ej. 1.3)} $\displaystyle \int (x + 1)(3x - 2)\,dx$ & = & $\displaystyle \int 3x^2 + x - 2\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont multiplicar factores y agrupar términos semejantes} \\[1mm]                                   & = & $\displaystyle 3 \int x^2\,dx + \int x\,dx - 2 \int dx$ & $\longleftarrow$     &  {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]                                   & = & $\displaystyle x^3 + \frac{x^2}{2} - 2x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}     \end{tabular}     \\[0.9cm]     \begin{tabular}{llllp{4cm}}     \textbf{Ej. 1.4)} $\displaystyle \int \sec y(\tan y - \sec y)\,dy$ & = & $\displaystyle \int \sec y \tan y - \sec^2y\,dy$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]                                    & $=$ & $\displaystyle \int \sec y \tan y\,dy - \int \sec^2 y\,dy$ &               $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[7mm]                                    & $=$ & $\displaystyle \sec y - \tan y + C$ & $\longleftarrow$  & {\fontsize{8}{0}\selectfont integrar}     \end{tabular}     \\[0.9cm]     \begin{tabular}{llllp{6cm}}     \textbf{Ej. 1.5)} $\displaystyle \int 2\pi y(8 - y^{3/2})\,dy$ & = & $\displaystyle 2\pi \int 8y - y^{5/2}\,dy$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont extraer el factor $2\pi$ fuera de la integral como una constante y reescribir la función}       \\[3mm]                     & = & $\displaystyle 2\pi \left[4y^2 - \frac{y^{7/2}}{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}    \\[6mm]                     & = & $\displaystyle 2\pi \left[4y^2 - \frac{2}{7}\, y^{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}     \end{tabular}     \\     \vfill     \tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]     {\hspace*{-0.55cm} {\Huge {\bfseries\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{El lector observará conforme vea los ejercicios elaborados de este texto que algunos de los pasos efectuados en los ejemplos 1.1 al 1.5 en la práctica son omitidos, esto ocurrirá a medida que se familiarice con las reglas básicas de integración.}     \newpage     Calcular las siguientes integrales     \\[0.55cm]     \hspace*{-0.25cm}     %----------------------------------------------------------     %                     LISTA DE EJERCICIOS     %----------------------------------------------------------     ${\setlength{\arraycolsep}{10pt} \begin{array}{*3{>{\displaystyle}l}} \textbf{1.6)} \int 2x - 3x^2\,dx & \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx & \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx \\[6mm] \textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx & \textbf{1.10)} \int \sqrt[3]{x^2}\,dx & \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx \\[6mm] \textbf{1.12)} \int (2t^2 - 1)^2\,dt & \textbf{1.13)} \int y^2\sqrt{y}\,dy & \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx \\[6mm] \textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt & \textbf{1.16)} \int \tan^2y + 1\,dy & \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx \\[6mm] \textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx & \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx & \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx \\[6mm] \textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx & \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx & \textbf{1.23)} \int \frac{x^2}{x^2 + 1}\,dx \\[6mm] \textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx & \textbf{1.25)} \int \tan^2x\,dx & \textbf{1.26)} \int \cot^2x\,dx \\[6mm] \textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}\,dx & \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx & \textbf{1.29)} \int (2^x + 3^x)^2\,dx \\[6mm] \textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx & \textbf{1.31)} \int 3^xe^x\,dx & \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx \\[6mm] \textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx & \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx & \textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy \\[6mm] \textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx & \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta & \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx \\[6mm] \textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx & \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} & \textbf{1.41)} \int \frac{dx}{3x^2 + 5} \\[6mm] \textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx & \textbf{1.43)} \int \left(y^2 - \frac{1}{y^2} \right)^3 dy & \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx \\[6mm] \textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx & \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt & \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}} \\[6mm] \textbf{1.48)} \int \frac{dx}{\sen x \cos x} & \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta & \textbf{1.50)} \int (\tan x + \sec x)^2\,dx \\[6mm] \textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx & \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx & \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi \\[6mm] \textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx & \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx & \textbf{1.56)} \int 1^x\,dx \end{array}}$     \newpage     \hspace*{-0.35cm}     ${\setlength{\arraycolsep}{10pt} \begin{array}{*3{>{\displaystyle}l}} \textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx & \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx & \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx \\[6mm] \textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx & \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx & \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx \\[6mm] \textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx & \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx & \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx \end{array}}$     \\[1.5cm]     \textbf{\huge Solución}     \\     \rule{21cm}{1ex}     \\[1ex]      %-------------------------------------------------------------     %                           EJERCICIO 1.6     %-------------------------------------------------------------      \begin{align*}     \textbf{1.6)} \int 2x - 3x^2\,dx &= \int 2x\,dx - \int 3x^2\,dx = 2 \int x\,dx - 3 \int x^2\,dx = \cancel{2}\left(\frac{x^2}{\cancel{2}} \right) - \cancel{3} \left(\frac{x^3}{\cancel{3}} \right) + C \\[3mm]                        &=  \fboxsep=5pt\colorbox{gris!40}{$x^2 - x^3 + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.7     %-------------------------------------------------------------      \begin{align*}     \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx &= \int 4x^3\,dx + \int 6x^2\,dx - \int dx                                                                                           = 4 \int x^3\,dx + 6 \int x^2\,dx - \int dx                                            \\[3mm]                                            &= \cancel{4} \left(\frac{x^4}{\cancel{4}} \right) + 6 \left( \frac{x^3}{3} \right) - x + C  =  \fboxsep=5pt\colorbox{gris!40}{$x^4 + 2x^3 - x + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.8     %-------------------------------------------------------------      \begin{align*}     \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx &= \int x^{3/2}\,dx + 2 \int x\,dx + \int dx                                                                                     = \frac{x^{5/2}}{5/2} + \cancel{2} \left(\frac{x^2}{\cancel{2}} \right) + x + C                                                                                     \\[3mm]                                                             &=  \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{5}\,x^{5/2} + x^2 + x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.9     %-------------------------------------------------------------      \begin{align*}     \textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx &= \int \sqrt{x}\,dx + \int \frac{dx}{2\sqrt{x}} = \int x^{1/2}\,dx + \frac{1}{2} \int x^{-1/2}\,dx = \frac{x^{3/2}}{3/2} + \frac{1}{\cancel{2}} \left(\frac{x^{1/2}}{1/ \cancel{2}} \right) + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{3}\,x^{3/2} + x^{1/2} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.10     %-------------------------------------------------------------       \begin{align*}     \textbf{1.10)} \int \sqrt[3]{x^2}\,dx = \int x^{2/3}\,dx = \frac{x^{5/3}}{5/3} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{5}\,x^{5/3} + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.11     %-------------------------------------------------------------       \begin{align*}     \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx &= \int \frac{x^2}{x^4}\,dx + 2 \int \frac{x}{x^4}\,dx - 3 \int \frac{dx}{x^4} = \int x^{-2}\,dx + 2 \int x^{-3}\,dx - 3 \int x^{-4}\,dx                   \end{align*}     \begin{align*}     &= \frac{x^{-1}}{- 1} + \cancel{2} \left(\frac{x^{-2}}{- \cancel{2}} \right) - \cancel{3} \left(\frac{x^{-3}}{- \cancel{3}} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.12     %-------------------------------------------------------------      \begin{align*}     \textbf{1.12)} \int (2t^2 - 1)^2\,dt &= \int 4t^4 - 4t^2 + 1\,dt = 4 \int t^4\,dt - 4 \int t^2\,dt + \int dt = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{4}{5} \, t^5 - \frac{4}{3} \, t^3 + t + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.13     %-------------------------------------------------------------      \begin{align*}     \textbf{1.13)} \int y^2\sqrt{y}\,dy = \int y^{5/2}\,dy = \frac{y^{7/2}}{7/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{7}\,y^{7/2} + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.14     %-------------------------------------------------------------      \begin{align*}     \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx = 2 \int \sen x\,dx + 3 \int \cos x\,dx = \fboxsep=5pt\colorbox{gris!40}{$- 2 \cos x + 3 \sen x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.15     %-------------------------------------------------------------      \begin{align*}     \textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt &= \int \frac{1 - t(t^2 + 1)}{t^2 + 1} = \int \frac{dt}{t^2 + 1} - \int \frac{t (\cancel{t^2 + 1})}{\cancel{t^2 + 1}}\,dt = \fboxsep=5pt\colorbox{gris!40}{$\arctan t - \displaystyle \frac{t^2}{2} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.16     %-------------------------------------------------------------      \begin{align*}     \textbf{1.16)} \int \tan^2y + 1\,dy = \int \sec^2y - 1 + 1\,dy = \fboxsep=5pt\colorbox{gris!40}{$\tan y + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.17     %-------------------------------------------------------------      \begin{align*}     \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx &= \int x^2\,dx + \int \frac{dx}{9x^2}                                          = \int x^2\,dx + \frac{1}{9} \int x^{-2}\,dx                                                                = \frac{x^3}{3} + \frac{1}{9}\left(\frac{x^{-1}}{-1}                                                                    \right) + C \\[3mm]                                &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{x^3}{3} - \frac{1}{9x} + C$}       \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.18     %-------------------------------------------------------------       \begin{align*}     \textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx &= \int \left(1 - \frac{1}{x^2} \right) \left(x^{3/2} \right)^{1/2}\,dx = \int \left(1 - \frac{1}{x^2} \right)x^{3/4}\,dx \\[3mm]     &= \int x^{3/4}\,dx - \int x^{-5/4}\,dx = \frac{x^{7/4}}{7/4} - \left(- \frac{x^{-1/4}}{1/4} \right) + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{4}{7}\,x^{7/4} + \frac{4}{x^{1/4}} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.19     %-------------------------------------------------------------      \begin{align*}     \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx &= \int \frac{2x - 2\sqrt{2x}\sqrt[3]{3x} + {(\sqrt[3]{3x})}^2}{x}\,dx \\[3mm]     &= 2 \int dx - 2 \int \frac{\sqrt{2}\,x^{1/2}\,\sqrt[3]{3}\,{x}^{1/3}}{x}\,dx + \int \frac{\sqrt[3]{9}\,x^{2/3}}{x}\,dx \\[3mm]     &= 2 \int dx - 2 \sqrt{2} \sqrt[3]{3} \int x^{-1/6}\,dx + \sqrt[3]{9} \int x^{-1/3}\,dx \\[3mm]     &= 2x - 2 \sqrt{2} \sqrt[3]{3} \left(\frac{x^{5/6}}{5/6} \right) + \sqrt[3]{9} \left( \frac{x^{2/3}}{2/3} \right) + C     \end{align*}     \begin{align*}     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2x - \frac{12}{5} \sqrt{2} \sqrt[3]{3}\,x^{5/6} + \frac{3}{2} \sqrt[3]{9}\,x^{2/3} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.20     %-------------------------------------------------------------      \begin{align*}     \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx &= \int \sqrt[3]{\sqrt{\frac{2x^2}{x}}}\,dx = \int \sqrt[3]{\sqrt{2x}}\,dx  = \int {\left[(2x)^{1/2} \right]}^{1/3}\,dx = \int (2x)^{1/6}\,dx  \\[3mm]     &= \sqrt[6]{2} \left(\frac{x^{7/6}}{7/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt[6]{2}}{7}\,x^{7/6} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.21     %-------------------------------------------------------------      \begin{align*}     \textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx &= \int \frac{2^x\,2}{10^x}\,dx - \int \frac{5^x}{5\,10^x}\,dx = 2 \int \left(\frac{2}{10} \right)^x\,dx - \frac{1}{5} \int \left(\frac{5}{10} \right)^x\,dx \\[3mm]     &= 2 \left[ \frac{(1/5)^x}{\ln(1/5)} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln(1/2)} \right] + C = 2 \left[ \frac{(1/5)^x}{\ln 1 - \ln 5} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln 1- \ln 2} \right] + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{5 \ln 2} \left(\frac{1}{2} \right)^x - \frac{2}{\ln 5} \left(\frac{1}{5} \right)^x + C$}     \end{align*}     %\reversemarginpar     \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Es menester destacar que en el ejercicio 1.23 no era necesario aplicar el algoritmo de la división, simplemente con sumar y restar el factor 1 en el numerador y separar las fracciones se obtendría el mismo resultado, este artificio de sumar y restar, multiplicar y dividir elementos será de gran utilidad para la resolución de un gran número de ejercicios presentados en este texto.     \vspace{-8pt}     \end{spacing}     \end{minipage}}     }      %-------------------------------------------------------------     %                          EJERCICIO 1.22     %-------------------------------------------------------------      \begin{align*}     \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx &= \int \frac{\sqrt{x^4 + \displaystyle \frac{1}{x^4} + 2}}{x^3}\,dx = \int \frac{\displaystyle \sqrt{\frac{x^8 + 2x^4 + 1}{x^4}}}{x^3}\,dx \\[3mm]     &= \int \frac{\sqrt{x^8 + 2x^4 + 1}}{x^5}\,dx = \int \frac{\sqrt{(x^4 + 1)^2}}{x^5}\,dx = \int \frac{x^4 + 1}{x^5}\,dx  \\[3mm]     &= \int \frac{dx}{x} + \int x^{-5}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \ln \big| x \big| - \frac{1}{4x^4} + C$}     \end{align*}     \vspace{1cm}       %-------------------------------------------------------------     %                          EJERCICIO 1.23     %-------------------------------------------------------------       \textbf{1.23)} $\displaystyle \int \frac{x^2}{x^2 + 1}\,dx$ \quad al aplicar la división de polinomios     \quad     $\begin{array}{cccc|ccc} & \cancel{x^2} & + & 0 & x^2 & + & 1 \\ \cline{5-7} - & \cancel{x^2} & - & 1 & 1 & & \\ \cline{2-4} & & - & 1 & & & \end{array}$     \\[0.25cm]     \begin{align*}     \mbox{La integral se convierte en} \int \frac{x^2}{x^2 + 1}\,dx &= \int \frac{(x^2 + 1)1 - 1}{x^2 + 1}\,dx = \int \frac{\cancel{x^2 + 1}}{\cancel{x^2 + 1}}\,dx - \int \frac{dx}{x^2 + 1}       \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle x - \arctan x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.24     %-------------------------------------------------------------      \begin{align*}     \textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx &= \int \frac{\cancel{(e^x + 1)}(e^{2x} - e^x + 1)}{\cancel{e^x + 1}}\,dx = \int e^{2x} - e^x + 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{e^{2x}}{2} - e^x + x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.25     %-------------------------------------------------------------      \begin{align*}     \textbf{1.25)} \int \tan^2x\,dx = \int \sec^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\tan x - x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.26     %-------------------------------------------------------------       \begin{align*}     \textbf{1.26)} \int \cot^2x\,dx = \int \csc^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$- \cot x - x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.27     %-------------------------------------------------------------       \begin{align*}     \textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx &= \int \frac{\sqrt{1 + x^2}}{\sqrt{1 - x^4}}\,dx + \int \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx \\[3mm]     &= \int \sqrt{\frac{\cancel{1 + x^2}}{(1 - x^2)\cancel{(1 + x^2)}}}\,dx + \int \sqrt{\frac{\cancel{1 - x^2}}{\cancel{(1 - x^2)}(1 + x^2)}}\,dx \\[3mm]     &= \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{dx}{\sqrt{x^2 + 1}} \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\arcsen x + \ln(x + \sqrt{x^2 + 1}) + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.28     %-------------------------------------------------------------      \begin{align*}     \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx &= \int \frac{1 - 3x^2 + 3x - x^3}{x\,x^{1/3}}\,dx     \\[3mm]     &= \int x^{-4/3}\,dx - 3 \int x^{2/3}\,dx + 3 \int x^{-1/3}\,dx - \int x^{5/3}\,dx \\[3mm]     &= \frac{x^{-1/3}}{-1/3} - 3 \, \frac{x^{5/3}}{5/3} + 3 \, \frac{x^{2/3}}{2/3} - \frac{x^{8/3}}{8/3} + C     \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{3}{x^{1/3}} - \frac{9}{5}\,x^{5/3} + \frac{9}{2}\,x^{2/3} - \frac{3}{8}\,x^{8/3} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.29     %-------------------------------------------------------------      \begin{align*}     \textbf{1.29)} \int (2^x + 3^x)^2\,dx &= \int {(2^x)}^2 + 2(2^x)(3^x) + {(3^x)}^2\,dx = \int 4^x\,dx + 2 \int 6^x\,dx + \int 9^x\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{\ln 4}\,4^x + \frac{2}{\ln 6}\,6^x + \frac{1}{\ln 9}\,9^x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.30     %-------------------------------------------------------------      \begin{align*}     \textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx &= \int n^{\frac{1 - n}{n}}\,x^{\frac{1 - n}{n}}\,dx     = n^{\frac{1 - n}{n}} \int x^{\frac{1 - n}{n}}\,dx = n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - n}{n} + 1}}{\frac{1 - n}{n} + 1} \right) + C \\[2mm]     &= n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - \cancel{n} + \cancel{n}}{n}}}{\frac{1 - \cancel{n} + \cancel{n}}{n}} \right) + C = n^{\frac{1 - n}{n}}\,n\,x^{1/n} + C = n^{1/n}\,x^{1/n} + C \\[2mm]     &= (nx)^{1/n} + C = \fboxsep=5pt\colorbox{gris!40}{$\sqrt[n]{nx} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.31     %-------------------------------------------------------------       \begin{align*}     \textbf{1.31)} \int 3^xe^x\,dx &= \int (3e)^x\,dx = \frac{(3e)^x}{\ln(3e)} + C = \frac{(3e)^x}{\ln 3 + \ln e} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3^xe^x}{\ln 3 + 1} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.32     %-------------------------------------------------------------       \begin{align*}     \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx &= \int {(a^{2/3})}^3 - 3{(a^{2/3})}^2x^{2/3} + 3a^{2/3}{(x^{2/3})}^2 - {(x^{2/3})}^3\,dx \\[3mm]     &= \int a^2 - 3a^{4/3}x^{2/3} + 3a^{2/3}x^{4/3} - x^2\,dx \\[3mm]     &= a^2 \int dx - 3a^{4/3} \int x^{2/3}\,dx + 3a^{2/3} \int x^{4/3}\,dx - \int x^2\,dx \\[3mm]     &= a^2x - 3a^{4/3}\left(\frac{x^{5/3}}{5/3} \right) + 3a^{2/3} \left(\frac{x^{7/3}}{7/3} \right) - \frac{x^3}{3} + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle a^2x - \frac{9a^{4/3}}{5}\,x^{5/3} + \frac{9a^{2/3}}{7}\,x^{7/3} - \frac{x^3}{3} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.33     %-------------------------------------------------------------      \begin{align*}     \textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx &= \int \frac{x^{2m} - 2x^mx^n + x^{2n}}{\sqrt{x}}\,dx  \\[3mm]     &= \int x^{2m - 1/2}\,dx - 2 \int x^{m + n - 1/2}\,dx + \int x^{2n - 1/2}\,dx \\[3mm]     &= \frac{x^{2m - 1/2 + 1}}{2m - 1/2 + 1} - 2 \left(\frac{x^{m + n - 1/2 + 1}}{m + n - 1/2 + 1} \right) + \frac{x^{2n - 1/2 + 1}}{2n - 1/2 + 1} + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2 \left( \frac{x^{\frac{4m + 1}{2}}}{4m + 1} \right) - 4 \left( \frac{x^{\frac{2m + 2n + 1}{2}}}{2m + 2n + 1} \right) + 2 \left( \frac{x^{\frac{4n + 1}{2}}}{4n + 1} \right) + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.34     %-------------------------------------------------------------      \begin{align*}     \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx &= \int \frac{a^{2x} - 2a^xb^x + b^{2x}}{a^xb^x}\,dx = \int \frac{a^{2x}}{a^xb^x}\,dx -2 \int \frac{\cancel{a^xb^x}}{\cancel{a^xb^x}}\,dx + \int \frac{b^{2x}}{a^xb^x}\,dx \\[2mm]     &= \int \left(\frac{a}{b} \right)^xdx - 2 \int dx + \int \left(\frac{b}{a} \right)^xdx \\[2mm]     &= \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln (a/b)} - 2x + \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln (b/a)} + C = \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln a - \ln b} - 2x - \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln a - \ln b} + C \\[2mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{\ln a - \ln b} \left[ \left(\frac{a}{b} \right)^x - \left(\frac{b}{a} \right)^x \right] - 2x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.35     %-------------------------------------------------------------       \begin{align*}     \textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy &= 4 \int y^3\,dy + 2 \int \frac{dy}{y^3} = \cancel{4}\left(\frac{y^4}{\cancel{4}} \right) - \cancel{2}\left(\frac{1}{- \cancel{2}y^2} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle y^4 + \frac{1}{y^2} + C$}     \end{align*}     \vspace{-0.65cm}       %-------------------------------------------------------------     %                          EJERCICIO 1.36     %-------------------------------------------------------------       \begin{align*}     \textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx &= \int \left(\frac{1}{\sqrt{2}\sen x} \right)^2 - 2\left(\frac{1}{\sqrt{2} \sen x} \right) + 1\,dx \\[3mm]     &= \int \frac{dx}{2 \sen^2x} - 2 \int \frac{dx}{\sqrt{2} \sen x} + \int dx = \frac{1}{2} \int \csc^2x\,dx - \sqrt{2} \int \csc x\,dx  + \int dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{2} \cot x - \sqrt{2} \ln \big|\csc x - \cot x \big| + x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.37     %-------------------------------------------------------------       \begin{align*}     \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta(1 + \tan^2\theta)}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta\,\cancel{\sec^2\theta}}{\cancel{\sec^2\theta}}\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \cos \theta + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.38     %-------------------------------------------------------------       \begin{align*}     \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx &= \int \frac{\sen x}{\cos^2x}\,dx = \int \frac{1}{\cos x}\,\frac{\sen x}{\cos x}\,dx = \int \sec \theta\,\tan \theta\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \sec \theta + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.39     %-------------------------------------------------------------       \begin{align*}     \textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx &= \int \frac{2\,\cancel{\sen x}\,\cos x}{\cancel{\sen x}}\,dx = \fboxsep=5pt\colorbox{gris!40}{$2\,\sen x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.40     %-------------------------------------------------------------       \begin{align*}     \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} &= \int \frac{dx}{(a - b) \left[\displaystyle \frac{a + b}{a - b} - x^2 \right]} = \frac{1}{a - b} \int \frac{dx}{\left(\sqrt{\frac{a + b}{a - b}} \right)^2 - x^2} \\[3mm]     &= \frac{1}{2(a - b)\displaystyle \sqrt{\frac{a + b}{a - b}}} \ln \left|\displaystyle \frac{\displaystyle \sqrt{\frac{a + b}{a - b}} + x}{\displaystyle \sqrt{\frac{a + b}{a - b}} - x} \right| + C \\[3mm]     &= \frac{1}{2\sqrt{(a + b)(a - b)}} \ln \left| \frac{\displaystyle  \frac{\sqrt{a + b} \ + \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}}{\displaystyle  \frac{\sqrt{a + b} \ - \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}} \right| + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{2 \sqrt{a^2 - b^2}} \ln \left| \frac{\sqrt{a + b} + \sqrt{a - b}x}{\sqrt{a + b} - \sqrt{a - b}x} \right| + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.41     %-------------------------------------------------------------        \begin{align*}     \textbf{1.41)} \int \frac{dx}{3x^2 + 5} &= \int \frac{dx}{3\left(x^2 + \frac{5}{3} \right)}     = \frac{1}{3} \int \frac{dx}{x^2 + \left(\sqrt{\frac{5}{3}} \right)^2} = \frac{1}{3} \left[\frac{1}{\sqrt{\frac{5}{3}}} \arctan \left(\frac{x}{\sqrt{\frac{5}{3}}} \right) \right] + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{\sqrt{3}}{3\sqrt{5}} \arctan \left(\frac{\sqrt{3} x}{\sqrt{5}} \right) + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.42     %-------------------------------------------------------------       \begin{align*}     \textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx &= \int {(\sqrt{x})}^2 - 2 \left(\cancel{\sqrt{x}} \frac{1}{\cancel{\sqrt{x}}} \right) + \left(\frac{1}{\sqrt{x}} \right)^2 dx = \int x^2\,dx - 2 \int dx + \int \frac{dx}{x} \\[3mm]     & = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{x^3}{3} - 2x + \ln \big| x \big| + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.43     %-------------------------------------------------------------       \begin{align*}     \textbf{1.43)}\int \left(y^2 - \frac{1}{y^2} \right)^3 dy &= \int {(y^2)}^3 - 3{(y^2)}^2\left(\frac{1}{y^2} \right) + 3y^2 \left(\frac{1}{y^2} \right)^2 - \left(\frac{1}{y^2} \right)^{\!\! 3} dy \\[3mm]     &= \int y^6\,dy - 3 \int y^2\,dy + 3 \int y^{-2}\,dy - \int y^{-6}\,dy \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{y^7}{7} - y^3 - \frac{3}{y} + \frac{1}{5y^5} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.44     %-------------------------------------------------------------       \begin{align*}     \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx &= \int \left( e^{x/a} \right)^2 - 2 e^{x/a} e^{- x/a} + \left( e^{- x/a} \right)^2 dx \\[3mm]     &= \int e^{2x/a} dx - 2 \int e^0\,dx + \int e^{-2x/a} dx \\[3mm]     &= \frac{e^{2x/a}}{\frac{2}{a}} - 2x - \frac{e^{- 2x/a}}{\frac{2}{a}} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{a}{2} \left[e^{2x/a} - e^{- 2x/a} \right] - 2x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.45     %-------------------------------------------------------------       \begin{align*}     \textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx &= \frac{\sqrt{5}}{5} \int x^{1/2}\,dx + \sqrt{5} \int x^{-1/2}\,dx = \frac{\sqrt{5}}{5} \left( \frac{2}{3} x^{3/2} \right) + \sqrt{5}(2 \sqrt{x}) + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2\sqrt{5}}{15}\,x^{3/2} + 2 \sqrt{5x} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.46     %-------------------------------------------------------------       \begin{align*}     \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt = 4 \int {(e^t)}^{-1/2} = - 4 \, \frac{e^{-t/2}}{1/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{8}{\sqrt{e^t}} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.47     %-------------------------------------------------------------       %\vspace*{-1cm}     \begin{align*}     \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}} &= \int \frac{dx}{\sqrt{5\left(\frac{7}{5} - x^2 \right)}} = \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(\sqrt{\frac{7}{5}} \right)^2 - x^2}} = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{\sqrt{5}}{5} \arcsen \left(\frac{\sqrt{5}x}{\sqrt{7}} \right) + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.48     %-------------------------------------------------------------       \begin{align*}     \textbf{1.48)} \int \frac{dx}{\sen x \cos x} &= \int \frac{\sen^2x + \cos^2x}{\sen x \cos x}\,dx = \int \frac{\sen^2x}{\sen x \cos x}\,dx + \int \frac{\cos^2x}{\sen x \cos x}\,dx = \int \tan x\,dx + \\[3mm]     & \int \cot x\,dx = - \ln \big| \cos x \big| + \ln \big| \sen x \big| + C = \ln \left| \frac{\sen x}{\cos x} \right| + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \ln \big| \tan x \big | + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.49     %-------------------------------------------------------------       \begin{align*}     \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta = \int \frac{d \theta}{\cos^2\theta} + \int \frac{\cancel{\cos^2 \theta}}{\cancel{\cos^2 \theta}}\,d \theta = \int \sec^2 \theta\,d \theta + \int d \theta     = \fboxsep=5pt\colorbox{gris!40}{$\tan \theta + \theta + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.50     %-------------------------------------------------------------       \begin{align*}     \textbf{1.50)} \int (\tan x + \sec x)^2\,dx &= \int \tan^2x + 2 \tan x\,\sec x + \sec^2x\,dx     \\[3mm]     &= \int \sec^2x - 1 + 2 \tan x\,\sec x + \sec^2x\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2\tan x + 2\sec x - x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.51     %-------------------------------------------------------------       \begin{align*}     \textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx = \int \left(x^{3/2} \right)^{1/2}dx = \int x^{3/4}dx = \frac{x^{7/4}}{7/4} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{4}{7}\,x^{7/4} + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.52     %-------------------------------------------------------------       \begin{align*}     \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx &= \int \frac{(3x^3 - 2) \cancel{(3x^3 + 2)}}{\cancel{(3x^3 + 2)}}\,dx = \int 3x^3\,dx - \int 2\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{4}\,x^4 - 2x + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.53     %-------------------------------------------------------------       \begin{align*}     \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi &= \int \frac{\displaystyle \frac{1}{\sen \phi}}{\displaystyle \frac{1}{\sen \phi} - \sen \phi}\,d\phi = \int \frac{\displaystyle \frac{1}{\cancel{\sen \phi}}}{\displaystyle \frac{1 - \sen^2\phi}{\cancel{\sen \phi}}}\,d \phi = \int \frac{1}{\cos^2\phi}\,d\phi = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \tan \phi + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.54     %-------------------------------------------------------------       \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Recuerde que la expresión $\int \frac{\sen x}{\cos^2x}\,dx$ ya había aparecido antes en el ejercicio 1.38 por lo que se omitieron algunos detalles de solución.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}     \vspace*{-0.25cm}     \begin{align*}     \textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx &= \int \frac{\sen x}{\cos x}\,dx + \int \frac{\tan x}{\cos x}\,dx = \int \tan x\,dx + \int \frac{\sen x}{\cos^2x}\,dx \\[3mm]     &= \int \tan x\,dx + \int \sec x \tan x\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \ln \big| \cos x \big| + \sec x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.55     %-------------------------------------------------------------      \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont La integral del ejercicio 1.56 es de la forma $a^x$ pero $a = 1$ por lo que no se puede aplicar la forma de la tabla, pero si las propiedades de los logaritmos.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}       \begin{align*}     \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx &= \frac{\sqrt{x(x - 1)} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx \\[3mm]     &= \int \frac{\sqrt{x}\sqrt{x - 1} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx = \int \frac{\cancel{\sqrt{x - 1}}(\sqrt{x} - e^x)}{\cancel{\sqrt{x - 1}}}\,dx \\[3mm]     &= \int \sqrt{x}\,dx - \int e^x\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{3}\,x^{3/2} - e^x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.56     %-------------------------------------------------------------       \begin{align*}     \textbf{1.56)} \int 1^x\,dx = \int e^{\ln 1^x}dx = \int e^{x\,\ln 1}dx = \int e^0dx = \int dx = \fboxsep=5pt\colorbox{gris!40}{$x + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.57     %-------------------------------------------------------------       \begin{align*}     \textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx &= \int x^{5/3}x^{-4/3}(x^3 - 1)\,dx = \int x^{1/3}(x^3 - 1)\,dx  \\[3mm]     &= \int x^{10/3}dx - \int x^{1/3}dx = \frac{x^{13/3}}{13/3} + \frac{x^{4/3}}{4/3} + C     \end{align*}     \begin{align*}     = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{13}\,x^{13/3} + \frac{3}{4}\,x^{4/3} + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.58     %-------------------------------------------------------------       \begin{align*}     \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx &= \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{(\sqrt{2x} - \sqrt{x})(\sqrt{2x} + \sqrt{x})}\,dx = \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{2x - x}\,dx     \\[3mm]     &= \int \frac{\sqrt{2}x^{3/2} + x^{3/2} - \sqrt{2}x^{1/2} - x^{1/2}}{x}\,dx \\[3mm]     &= \sqrt{2} \int x^{1/2}\,dx + \int x^{1/2}\,dx - \sqrt{2} \int x^{-1/2}\,dx - \int x^{- 1/2}\,dx \\[3mm]     &= ( \sqrt{2} + 1) \int x^{1/2}\,dx - (\sqrt{2} + 1) \int x^{-1/2}\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle (\sqrt{2} + 1) \left[\frac{2}{3}\,x^{3/2} - 2x^{1/2} \right] + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.59     %-------------------------------------------------------------       \begin{align*}     \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx &= \frac{\sqrt{5}}{\sqrt[3]{3}} \int \frac{x^{3/2}}{x^{1/3}}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \int x^{7/6}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \left( \frac{x^{13/6}}{13/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt{5}}{13 \sqrt[3]{3}}\,x^{13/6} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.60     %-------------------------------------------------------------       \begin{align*}     \textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx &= \int \frac{\tan x}{3 \sen x}\,dx - \int \frac{\sen^2x}{3 \sen x}\,dx + \int \frac{4 \cos x}{3 \sen x}\,dx \\[3mm]     &= \frac{1}{3} \int \frac{\cancel{\sen x}}{\cos x \, \cancel{\sen x}}\,dx - \frac{1}{3} \int \sen x\,dx + \frac{4}{3} \int \cot x\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{3}\,\ln \big| \sec x + \tan x \big| + \frac{1}{3} \cos x + \frac{4}{3} \ln \big| \sen x \big| + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.61     %-------------------------------------------------------------       \begin{align*}     \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx &= \int \left(\frac{1}{x} \right)^3 - 3 \left(\frac{1}{x} \right)^2 x + 3 \left(\frac{1}{x} \right)\,x^2 - x^3\,dx \\[3mm]     &= \int x^{-3}\,dx - 3 \int \frac{dx}{x} + 3 \int x\,dx - \int x^3\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{2\,x^2} - 3\,\ln \big| x \big| + \frac{3}{2}\,x^2 - \frac{x^4}{4} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.62     %-------------------------------------------------------------       \begin{align*}     \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx = \int \frac{\cancel{e^x}\,e^2}{\cancel{e^x}\,e}\,dx = \int e\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle ex + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.63     %-------------------------------------------------------------       \begin{align*}     \textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx = \int 8x^3 - 6x^2 - x^{-3}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2x^4 - 2x^3 + \frac{1}{2x^2} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.64     %-------------------------------------------------------------       \begin{align*}     \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx = \int \frac{4 \, \cancel{\ln x}}{\cancel{\ln x}}\,dx = 4 \int dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 4x + C$}     \end{align*}     \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Observe con cuidado que en el ejercicio 1.64, la función logaritmo del numerador \emph{no está elevada a la cuarta potencia}, solo su argumento, por eso fue posible la aplicación de la propiedad y la posterior la simplificación de los términos.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}       %-------------------------------------------------------------     %                          EJERCICIO 1.65     %-------------------------------------------------------------       \vspace*{-0.8cm}     \begin{align*}     \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx &= \int \frac{1 - \sen x}{(1 + \sen x)(1 - \sen x)}\,dx = \int \frac{1 - \sen x}{1 - \sen^2x}\,dx     = \int \frac{1 - \sen x}{\cos^2x}\,dx     \\[3mm]     &= \int \sec^2x\,dx - \int \frac{\sen x}{\cos^2x}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \tan x - \sec x + C$}     \end{align*}     \end{document}

has the problem in the picture
Attachments
math-error.png (38.71 KiB) Viewed 4770 times

Stefan Kottwitz
Posts: 9550
Joined: Mon Mar 10, 2008 9:44 pm
You seth \mathindent, that is for left indenting. But by default, equations are centered. Either remove the line 56

\setlength{\mathindent}{1pt}

or add fleqn as class option at the beginning.

Stefan

danielvelizv
Posts: 45
Joined: Wed May 25, 2016 7:04 am
Stefan_K wrote:You seth \mathindent, that is for left indenting. But by default, equations are centered. Either remove the line 56

\setlength{\mathindent}{1pt}

or add fleqn as class option at the beginning.

Stefan

Ok, but it didn't solve the main problem of this post, I changed this time to fleqn and \setlength{\mathindent}{0pt}, but it produces the same result of the image
Attachments
align-integral.png (18.39 KiB) Viewed 4759 times

Johannes_B
Site Moderator
Posts: 4109
Joined: Thu Nov 01, 2012 4:08 pm
You need both.
The smart way: Calm down and take a deep breath, read posts and provided links attentively, try to understand and ask if necessary.

danielvelizv
Posts: 45
Joined: Wed May 25, 2016 7:04 am
Johannes_B wrote:You need both.

I'm using both fleqn and \setlength{\mathindent}{0pt}, but no good results were achieved, help please! I compiled with LaTeX, the bf text must to be left aligned in the first line

Stefan helped me creating this code, but for some reason, it didn't produce the result I expected

\documentclass[10pt,letterpaper,fleqn]{scrbook}\usepackage[left=3cm,right=2.5cm,top=2.5cm,bottom=2.5cm,   marginparwidth=2.85cm, marginparsep=0pt]{geometry}\usepackage[leqno]{mathtools}\newtagform{bold}{}{\textbf{)}}\usetagform{bold}\renewcommand*{\theequation}{\textbf{Ej. 1.}}\newcommand{\remark}[1]{& \quad \longleftarrow   & \quad\fontsize{8}{0}\selectfont\parbox{0.25\textwidth}{\raggedright#1}}\setlength{\jot}{10pt}\begin{document}\begin{equation}\begin{alignedat}[t]{2}   \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx       &= \int \frac{x^2}{\sqrt{x}}\,dx          + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx          \remark{separar en 3 integrales} \\       &= \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx          + \int \frac{1}{x^{1/2}}\,dx          \remark{escribir los radicales en forma de potencia} \\       &= \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx          \remark{reescribir} \\       &= \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C          \remark{integrar} \\       &= \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} +          \remark{simplificar}     \end{alignedat}\end{equation}\end{document}
Attachments
align-integral.png (18.39 KiB) Viewed 4731 times

Stefan Kottwitz
Posts: 9550
Joined: Mon Mar 10, 2008 9:44 pm
This code produces the result with equation number at the left side. Not your screenshot.
Just post your code that gives the wrong result, and not my original code that gives the correct result, so there's nothing to correct. No fault here. Just show your code that belongs to the output.

Stefan

danielvelizv
Posts: 45
Joined: Wed May 25, 2016 7:04 am
Stefan_K wrote:This code produces the result with equation number at the left side. Not your screenshot.
Just post your code that gives the wrong result, and not my original code that gives the correct result, so there's nothing to correct. No fault here. Just show your code that belongs to the output.

Stefan

If it is not too much trouble, here is the code where I have the problem and I want to correct, I will need help to make macros in some things like the array where I have written the numbered integrals, where can I read about macros in LaTeX? Regards!

     \documentclass[10pt,letterpaper,fleqn,     headsepline,footsepline,     plainheadsepline,plainfootsepline,     ]{scrbook}     \usepackage[left=3cm,right=2.5cm,top=2.5cm,bottom=2.5cm, marginparwidth=2.85cm, marginparsep=0pt,head=22.22223pt]{geometry}     \usepackage[utf8]{inputenc}     \usepackage[spanish]{babel}     \usepackage{amsmath}     \usepackage{mathtools}            %MANIPULACIÓN DE LA ALINEACIÓN LATERAL DE LAS EXPRESIONES MATEMÁTICAS     \usepackage{amsfonts}     \usepackage{amssymb}     \usepackage{graphicx}     \usepackage[most]{tcolorbox}     \usepackage{xcolor}     \usepackage{tikz}     \usepackage{array}     \usepackage{marginnote}           %COLOCACIÓN DE NOTAS DE PÁGINA EN LOS LADOS     \usepackage{setspace}             %SEPARACIÓN DE LÍNEAS EN PÁRRAFOS     \usepackage{cancel}               %CANCELACIÓN DE TÉRMINOS     \usetikzlibrary{calc}     \usetikzlibrary{shapes.callouts}  %CUADROS DE IDEAS     \usetikzlibrary{decorations.text}     \usetikzlibrary{positioning}     \usepackage{varwidth}     \def\cabecera#1{       \begin{tikzpicture}[overlay, remember picture]         \draw let \p1 = (current page.west), \p2 = (current page.east) in           node[minimum width=\x2-\x1, minimum height=3cm, line width=0pt, rectangle, fill=gris!80, anchor=north west, align=left, text width=17cm] at ($(current page.north west)$)    {#1};       \end{tikzpicture}     } 	\newtagform{bold}{}{\textbf{)}}	\usetagform{bold}	\renewcommand*{\theequation}{\textbf{Ej. 1.}}	\newcommand{\remark}[1]{& \quad \longleftarrow   & \quad\fontsize{8}{0}\selectfont\parbox{0.25\textwidth}{\raggedright#1}}	\setlength{\jot}{10pt}      \newcommand{\titulo}{{650 integrales indefinidas \\ resueltas ¡paso a paso!}}     \newcommand{\inmediata}{{Integrales inmediatas}}     \usepackage{scrlayer-scrpage}     \addtokomafont{pagenumber}{\bfseries}     \addtokomafont{pageheadfoot}{\fontsize{8}{9}\sffamily\upshape}     \clearpairofpagestyles     \ohead*{\inmediata}     \ihead*{\titulo}     \ofoot*{\pagemark}     \ifoot*{Ing. Daniel A. Veliz V.}        \setlength{\parindent}{0pt}      %SIN SANGRÍA EN LOS PÁRRAFOS     \setlength{\mathindent}{0pt}     %SIN SANGRÍA EN LAS EXPRESIONES MATEMÁTICAS     \setlength{\arraycolsep}{4pt}    %ANCHO DE LAS COLUMNAS EN LOS ARRAY     \setlength{\tabcolsep}{4pt}      %ANCHO DE LAS COLUMNAS EN LAS TABLAS     \usepackage{anyfontsize}       %---------------------------------------------     %              COLORES DEFINIDOS     %---------------------------------------------       \definecolor{naranja}{rgb}{1, 0.3, 0}     \definecolor{blanco}{rgb}{0.97, 0.97, 1}     \definecolor{gris}{rgb}{0.47, 0.53, 0.6}     \definecolor{azul}{rgb}{0.12, 0.56, 1.0}     \definecolor{verde}{rgb}{0.0, 0.65, 0.31}     \definecolor{carmin}{rgb}{1.0, 0.0, 0.22}       \begin{document}     \cabecera{\bfseries {\fontsize{20}{0}\selectfont  \hfill Capítulo I \\ \hfill Integrales inmediatas}}     \begin{minipage}[c]{1\textwidth}     \vspace*{1cm}     En este capítulo se darán a conocer los fundamentos básicos de la integración de distintas funciones por medio del empleo de las propiedades matemáticas y así convertir las funciones integrando dadas en algunas de las formas básicas presentadas antes del desarrollo de este capítulo, de esta manera a medida que revise los capitulos posteriores se dará cuenta que la idea básica de aplicar las técnicas de integración consistirá en convertir integrandos complicados en formas elementales para determinar una \emph{función primitiva} o \emph{antiderivada} de una función $f$.     \\[0.5cm]     La antiderivación (o integración indefinida) se denota mediante el signo integral     $\displaystyle \int$ por lo tanto, el siguiente esquema podrá ayudarlo a identificar los elementos implícitos en el cálculo integral y qué se obtiene al calcular una integral indefinida:     \end{minipage}     \\[0.8cm]     \hspace*{3.75cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=azul!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0.7,-1)}, callout pointer width=5mm] {\begin{varwidth}{2cm} Función integrando \end{varwidth}};     \end{tikzpicture}     \hspace*{1.5cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=verde!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.5,-1)}, callout pointer width=5mm] {\begin{varwidth}{2.75cm} Antiderivada de la función $f$ \end{varwidth}};     \end{tikzpicture}     \\[-0.675cm]     \begin{equation}     \hspace*{5cm} \scalebox{1.5}{$\displaystyle \int \! \textcolor{azul!100}{f(x)}\,\textcolor{naranja!90}{dx} = \textcolor{verde!100}{F(x)} \ \textcolor{carmin!100}{+ \ C}$} \nonumber     \end{equation}     \\[-0.6cm]     \hspace*{5.85cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=naranja!90, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0,1)}, callout pointer width=5mm] {\begin{varwidth}{2cm}     Variable de integración \end{varwidth}};     \end{tikzpicture}     \hspace*{1.75cm}     \begin{tikzpicture}     \node[rectangle callout, rounded corners=3pt, draw, fill=carmin!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.75,1)}, callout pointer width=5mm] {\begin{varwidth}{2.5cm}     Constante de integración \end{varwidth}};     \end{tikzpicture}     \\[0.75cm]     Además, según Larson R. (2009) en su texto \emph{Cálculo Integral - Matemáticas 2} expresa que:     \vspace{1ex}     \begin{quote}     La expresión $\displaystyle \int f(x)\,dx$ se lee como la antiderivada o primitiva de $f$ con respecto a $x$, el diferencial de $x$ sirve para identificar a $x$ como la variable de integración. El término \emph{integral indefinida} es sinónimo de antiderivada."     \end{quote}     \vspace{0.5cm}     \tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]     {\hspace*{-0.55cm} {\Huge {\bfseries\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{A lo largo de este texto, usted encontrará conforme vea las distintas técnicas y casos de integrandos particulares, la complejidad en el desarrollo de los mismos, como integrar cada función producirá una constante $C$, solo se asumirá en el resultado final escrito como la suma de todas las constantes de las integrales resueltas, de manera que $C = C_1 + C_2 + C_3 + \ldots + C_n$}     \vspace{0.6cm}     A continuación se presentará una lista de ejercicios con un orden aleatorio de dificultad y algunos ejemplos previamente explicados para ayudar a comprender el principio básico de la integración inmediata por medio del uso de la tabla.     \\[0.6cm]\begin{equation}\begin{alignedat}[t]{2}   \int x + 3\,dx       &= \int x\,dx + \int 3\,dx          \remark{separar en 2 integrales} \\       &= \int x\,dx + 3 \int dx          \remark{en la integral de la derecha se extrajo el factor 3 fuera de la integral como una constante} \\       &= \frac{x^2}{2} + 3x + C          \remark{integrar}     \end{alignedat}\end{equation}    \begin{equation}\begin{alignedat}[t]{2}\int \frac{x^2 + x + 1}{\sqrt{x}}\,dx 	  & = \int \frac{x^2}{\sqrt{x}}\,dx + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx	  \remark{separar en 3 integrales} \\	  & = \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx + \int \frac{1}{x^{1/2}}\,dx	  \remark{escribir los radicales en forma de potencia} \\	  & = \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx	  \remark{reescribir}  \\	  & = \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C	  \remark{integrar} \\	  & = \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} + C	  \remark{simplificar} \end{alignedat}\end{equation}%    \begin{tabular}{llllp{3cm}}     %\textbf{Ej. 1.2)} $\displaystyle \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx$ & = & $\displaystyle \int \frac{x^2}{\sqrt{x}}\,dx + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]     %                              & = & $\displaystyle \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx + \int \frac{1}{x^{1/2}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont escribir los radicales en forma de potencia}    \\[1mm]     %                                & = & $\displaystyle \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]     %                                & = & $\displaystyle \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar} \\[6mm]     %                               & = & $\displaystyle \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}     %\end{tabular}     \reversemarginpar     \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Las funciones irracionales (raíces) cuentan como funciones de potencia.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}     \\[0.9cm]     \begin{tabular}{llllp{4cm}}     \textbf{Ej. 1.3)} $\displaystyle \int (x + 1)(3x - 2)\,dx$ & = & $\displaystyle \int 3x^2 + x - 2\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont multiplicar factores y agrupar términos semejantes} \\[1mm]                                   & = & $\displaystyle 3 \int x^2\,dx + \int x\,dx - 2 \int dx$ & $\longleftarrow$     &  {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]                                   & = & $\displaystyle x^3 + \frac{x^2}{2} - 2x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}     \end{tabular}     \\[0.9cm]     \begin{tabular}{llllp{4cm}}     \textbf{Ej. 1.4)} $\displaystyle \int \sec y(\tan y - \sec y)\,dy$ & = & $\displaystyle \int \sec y \tan y - \sec^2y\,dy$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]                                    & $=$ & $\displaystyle \int \sec y \tan y\,dy - \int \sec^2 y\,dy$ &               $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[7mm]                                    & $=$ & $\displaystyle \sec y - \tan y + C$ & $\longleftarrow$  & {\fontsize{8}{0}\selectfont integrar}     \end{tabular}     \\[0.9cm]     \begin{tabular}{llllp{6cm}}     \textbf{Ej. 1.5)} $\displaystyle \int 2\pi y(8 - y^{3/2})\,dy$ & = & $\displaystyle 2\pi \int 8y - y^{5/2}\,dy$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont extraer el factor $2\pi$ fuera de la integral como una constante y reescribir la función}       \\[3mm]                     & = & $\displaystyle 2\pi \left[4y^2 - \frac{y^{7/2}}{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}    \\[6mm]                     & = & $\displaystyle 2\pi \left[4y^2 - \frac{2}{7}\, y^{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}     \end{tabular}     \\     \vfill     \tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]     {\hspace*{-0.55cm} {\Huge {\bfseries\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{El lector observará conforme vea los ejercicios elaborados de este texto que algunos de los pasos efectuados en los ejemplos 1.1 al 1.5 en la práctica son omitidos, esto ocurrirá a medida que se familiarice con las reglas básicas de integración.}     \newpage     Calcular las siguientes integrales     \\[0.55cm]     \hspace*{-0.25cm}     %----------------------------------------------------------     %                     LISTA DE EJERCICIOS     %----------------------------------------------------------     ${\setlength{\arraycolsep}{10pt} \begin{array}{*3{>{\displaystyle}l}} \textbf{1.6)} \int 2x - 3x^2\,dx & \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx & \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx \\[6mm] \textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx & \textbf{1.10)} \int \sqrt[3]{x^2}\,dx & \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx \\[6mm] \textbf{1.12)} \int (2t^2 - 1)^2\,dt & \textbf{1.13)} \int y^2\sqrt{y}\,dy & \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx \\[6mm] \textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt & \textbf{1.16)} \int \tan^2y + 1\,dy & \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx \\[6mm] \textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx & \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx & \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx \\[6mm] \textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx & \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx & \textbf{1.23)} \int \frac{x^2}{x^2 + 1}\,dx \\[6mm] \textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx & \textbf{1.25)} \int \tan^2x\,dx & \textbf{1.26)} \int \cot^2x\,dx \\[6mm] \textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}\,dx & \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx & \textbf{1.29)} \int (2^x + 3^x)^2\,dx \\[6mm] \textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx & \textbf{1.31)} \int 3^xe^x\,dx & \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx \\[6mm] \textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx & \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx & \textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy \\[6mm] \textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx & \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta & \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx \\[6mm] \textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx & \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} & \textbf{1.41)} \int \frac{dx}{3x^2 + 5} \\[6mm] \textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx & \textbf{1.43)} \int \left(y^2 - \frac{1}{y^2} \right)^3 dy & \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx \\[6mm] \textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx & \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt & \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}} \\[6mm] \textbf{1.48)} \int \frac{dx}{\sen x \cos x} & \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta & \textbf{1.50)} \int (\tan x + \sec x)^2\,dx \\[6mm] \textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx & \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx & \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi \\[6mm] \textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx & \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx & \textbf{1.56)} \int 1^x\,dx \end{array}}$     \newpage     \hspace*{-0.35cm}     ${\setlength{\arraycolsep}{10pt} \begin{array}{*3{>{\displaystyle}l}} \textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx & \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx & \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx \\[6mm] \textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx & \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx & \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx \\[6mm] \textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx & \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx & \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx \end{array}}$     \\[1.5cm]     \textbf{\huge Solución}     \\     \rule{21cm}{1ex}     \\[1ex]      %-------------------------------------------------------------     %                           EJERCICIO 1.6     %-------------------------------------------------------------      \begin{align*}     \textbf{1.6)} \int 2x - 3x^2\,dx &= \int 2x\,dx - \int 3x^2\,dx = 2 \int x\,dx - 3 \int x^2\,dx = \cancel{2}\left(\frac{x^2}{\cancel{2}} \right) - \cancel{3} \left(\frac{x^3}{\cancel{3}} \right) + C \\[3mm]                        &=  \fboxsep=5pt\colorbox{gris!40}{$x^2 - x^3 + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.7     %-------------------------------------------------------------      \begin{align*}     \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx &= \int 4x^3\,dx + \int 6x^2\,dx - \int dx                                                                                           = 4 \int x^3\,dx + 6 \int x^2\,dx - \int dx                                            \\[3mm]                                            &= \cancel{4} \left(\frac{x^4}{\cancel{4}} \right) + 6 \left( \frac{x^3}{3} \right) - x + C  =  \fboxsep=5pt\colorbox{gris!40}{$x^4 + 2x^3 - x + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.8     %-------------------------------------------------------------      \begin{align*}     \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx &= \int x^{3/2}\,dx + 2 \int x\,dx + \int dx                                                                                     = \frac{x^{5/2}}{5/2} + \cancel{2} \left(\frac{x^2}{\cancel{2}} \right) + x + C                                                                                     \\[3mm]                                                             &=  \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{5}\,x^{5/2} + x^2 + x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.9     %-------------------------------------------------------------      \begin{align*}     \textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx &= \int \sqrt{x}\,dx + \int \frac{dx}{2\sqrt{x}} = \int x^{1/2}\,dx + \frac{1}{2} \int x^{-1/2}\,dx = \frac{x^{3/2}}{3/2} + \frac{1}{\cancel{2}} \left(\frac{x^{1/2}}{1/ \cancel{2}} \right) + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{3}\,x^{3/2} + x^{1/2} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.10     %-------------------------------------------------------------       \begin{align*}     \textbf{1.10)} \int \sqrt[3]{x^2}\,dx = \int x^{2/3}\,dx = \frac{x^{5/3}}{5/3} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{5}\,x^{5/3} + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.11     %-------------------------------------------------------------       \begin{align*}     \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx &= \int \frac{x^2}{x^4}\,dx + 2 \int \frac{x}{x^4}\,dx - 3 \int \frac{dx}{x^4} = \int x^{-2}\,dx + 2 \int x^{-3}\,dx - 3 \int x^{-4}\,dx                   \end{align*}     \begin{align*}     &= \frac{x^{-1}}{- 1} + \cancel{2} \left(\frac{x^{-2}}{- \cancel{2}} \right) - \cancel{3} \left(\frac{x^{-3}}{- \cancel{3}} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.12     %-------------------------------------------------------------      \begin{align*}     \textbf{1.12)} \int (2t^2 - 1)^2\,dt &= \int 4t^4 - 4t^2 + 1\,dt = 4 \int t^4\,dt - 4 \int t^2\,dt + \int dt = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{4}{5} \, t^5 - \frac{4}{3} \, t^3 + t + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.13     %-------------------------------------------------------------      \begin{align*}     \textbf{1.13)} \int y^2\sqrt{y}\,dy = \int y^{5/2}\,dy = \frac{y^{7/2}}{7/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{7}\,y^{7/2} + C$}       \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.14     %-------------------------------------------------------------      \begin{align*}     \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx = 2 \int \sen x\,dx + 3 \int \cos x\,dx = \fboxsep=5pt\colorbox{gris!40}{$- 2 \cos x + 3 \sen x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.15     %-------------------------------------------------------------      \begin{align*}     \textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt &= \int \frac{1 - t(t^2 + 1)}{t^2 + 1} = \int \frac{dt}{t^2 + 1} - \int \frac{t (\cancel{t^2 + 1})}{\cancel{t^2 + 1}}\,dt = \fboxsep=5pt\colorbox{gris!40}{$\arctan t - \displaystyle \frac{t^2}{2} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.16     %-------------------------------------------------------------      \begin{align*}     \textbf{1.16)} \int \tan^2y + 1\,dy = \int \sec^2y - 1 + 1\,dy = \fboxsep=5pt\colorbox{gris!40}{$\tan y + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.17     %-------------------------------------------------------------      \begin{align*}     \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx &= \int x^2\,dx + \int \frac{dx}{9x^2}                                          = \int x^2\,dx + \frac{1}{9} \int x^{-2}\,dx                                                                = \frac{x^3}{3} + \frac{1}{9}\left(\frac{x^{-1}}{-1}                                                                    \right) + C \\[3mm]                                &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{x^3}{3} - \frac{1}{9x} + C$}       \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.18     %-------------------------------------------------------------       \begin{align*}     \textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx &= \int \left(1 - \frac{1}{x^2} \right) \left(x^{3/2} \right)^{1/2}\,dx = \int \left(1 - \frac{1}{x^2} \right)x^{3/4}\,dx \\[3mm]     &= \int x^{3/4}\,dx - \int x^{-5/4}\,dx = \frac{x^{7/4}}{7/4} - \left(- \frac{x^{-1/4}}{1/4} \right) + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{4}{7}\,x^{7/4} + \frac{4}{x^{1/4}} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.19     %-------------------------------------------------------------      \begin{align*}     \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx &= \int \frac{2x - 2\sqrt{2x}\sqrt[3]{3x} + {(\sqrt[3]{3x})}^2}{x}\,dx \\[3mm]     &= 2 \int dx - 2 \int \frac{\sqrt{2}\,x^{1/2}\,\sqrt[3]{3}\,{x}^{1/3}}{x}\,dx + \int \frac{\sqrt[3]{9}\,x^{2/3}}{x}\,dx \\[3mm]     &= 2 \int dx - 2 \sqrt{2} \sqrt[3]{3} \int x^{-1/6}\,dx + \sqrt[3]{9} \int x^{-1/3}\,dx \\[3mm]     &= 2x - 2 \sqrt{2} \sqrt[3]{3} \left(\frac{x^{5/6}}{5/6} \right) + \sqrt[3]{9} \left( \frac{x^{2/3}}{2/3} \right) + C     \end{align*}     \begin{align*}     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2x - \frac{12}{5} \sqrt{2} \sqrt[3]{3}\,x^{5/6} + \frac{3}{2} \sqrt[3]{9}\,x^{2/3} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.20     %-------------------------------------------------------------      \begin{align*}     \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx &= \int \sqrt[3]{\sqrt{\frac{2x^2}{x}}}\,dx = \int \sqrt[3]{\sqrt{2x}}\,dx  = \int {\left[(2x)^{1/2} \right]}^{1/3}\,dx = \int (2x)^{1/6}\,dx  \\[3mm]     &= \sqrt[6]{2} \left(\frac{x^{7/6}}{7/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt[6]{2}}{7}\,x^{7/6} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.21     %-------------------------------------------------------------      \begin{align*}     \textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx &= \int \frac{2^x\,2}{10^x}\,dx - \int \frac{5^x}{5\,10^x}\,dx = 2 \int \left(\frac{2}{10} \right)^x\,dx - \frac{1}{5} \int \left(\frac{5}{10} \right)^x\,dx \\[3mm]     &= 2 \left[ \frac{(1/5)^x}{\ln(1/5)} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln(1/2)} \right] + C = 2 \left[ \frac{(1/5)^x}{\ln 1 - \ln 5} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln 1- \ln 2} \right] + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{5 \ln 2} \left(\frac{1}{2} \right)^x - \frac{2}{\ln 5} \left(\frac{1}{5} \right)^x + C$}     \end{align*}     %\reversemarginpar     \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Es menester destacar que en el ejercicio 1.23 no era necesario aplicar el algoritmo de la división, simplemente con sumar y restar el factor 1 en el numerador y separar las fracciones se obtendría el mismo resultado, este artificio de sumar y restar, multiplicar y dividir elementos será de gran utilidad para la resolución de un gran número de ejercicios presentados en este texto.     \vspace{-8pt}     \end{spacing}     \end{minipage}}     }      %-------------------------------------------------------------     %                          EJERCICIO 1.22     %-------------------------------------------------------------      \begin{align*}     \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx &= \int \frac{\sqrt{x^4 + \displaystyle \frac{1}{x^4} + 2}}{x^3}\,dx = \int \frac{\displaystyle \sqrt{\frac{x^8 + 2x^4 + 1}{x^4}}}{x^3}\,dx \\[3mm]     &= \int \frac{\sqrt{x^8 + 2x^4 + 1}}{x^5}\,dx = \int \frac{\sqrt{(x^4 + 1)^2}}{x^5}\,dx = \int \frac{x^4 + 1}{x^5}\,dx  \\[3mm]     &= \int \frac{dx}{x} + \int x^{-5}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \ln \big| x \big| - \frac{1}{4x^4} + C$}     \end{align*}     \vspace{1cm}       %-------------------------------------------------------------     %                          EJERCICIO 1.23     %-------------------------------------------------------------       \textbf{1.23)} $\displaystyle \int \frac{x^2}{x^2 + 1}\,dx$ \quad al aplicar la división de polinomios     \quad     $\begin{array}{cccc|ccc} & \cancel{x^2} & + & 0 & x^2 & + & 1 \\ \cline{5-7} - & \cancel{x^2} & - & 1 & 1 & & \\ \cline{2-4} & & - & 1 & & & \end{array}$     \\[0.25cm]     \begin{align*}     \mbox{La integral se convierte en} \int \frac{x^2}{x^2 + 1}\,dx &= \int \frac{(x^2 + 1)1 - 1}{x^2 + 1}\,dx = \int \frac{\cancel{x^2 + 1}}{\cancel{x^2 + 1}}\,dx - \int \frac{dx}{x^2 + 1}       \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle x - \arctan x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.24     %-------------------------------------------------------------      \begin{align*}     \textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx &= \int \frac{\cancel{(e^x + 1)}(e^{2x} - e^x + 1)}{\cancel{e^x + 1}}\,dx = \int e^{2x} - e^x + 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{e^{2x}}{2} - e^x + x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.25     %-------------------------------------------------------------      \begin{align*}     \textbf{1.25)} \int \tan^2x\,dx = \int \sec^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\tan x - x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.26     %-------------------------------------------------------------       \begin{align*}     \textbf{1.26)} \int \cot^2x\,dx = \int \csc^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$- \cot x - x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.27     %-------------------------------------------------------------       \begin{align*}     \textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx &= \int \frac{\sqrt{1 + x^2}}{\sqrt{1 - x^4}}\,dx + \int \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx \\[3mm]     &= \int \sqrt{\frac{\cancel{1 + x^2}}{(1 - x^2)\cancel{(1 + x^2)}}}\,dx + \int \sqrt{\frac{\cancel{1 - x^2}}{\cancel{(1 - x^2)}(1 + x^2)}}\,dx \\[3mm]     &= \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{dx}{\sqrt{x^2 + 1}} \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\arcsen x + \ln(x + \sqrt{x^2 + 1}) + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.28     %-------------------------------------------------------------      \begin{align*}     \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx &= \int \frac{1 - 3x^2 + 3x - x^3}{x\,x^{1/3}}\,dx     \\[3mm]     &= \int x^{-4/3}\,dx - 3 \int x^{2/3}\,dx + 3 \int x^{-1/3}\,dx - \int x^{5/3}\,dx \\[3mm]     &= \frac{x^{-1/3}}{-1/3} - 3 \, \frac{x^{5/3}}{5/3} + 3 \, \frac{x^{2/3}}{2/3} - \frac{x^{8/3}}{8/3} + C     \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{3}{x^{1/3}} - \frac{9}{5}\,x^{5/3} + \frac{9}{2}\,x^{2/3} - \frac{3}{8}\,x^{8/3} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.29     %-------------------------------------------------------------      \begin{align*}     \textbf{1.29)} \int (2^x + 3^x)^2\,dx &= \int {(2^x)}^2 + 2(2^x)(3^x) + {(3^x)}^2\,dx = \int 4^x\,dx + 2 \int 6^x\,dx + \int 9^x\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{\ln 4}\,4^x + \frac{2}{\ln 6}\,6^x + \frac{1}{\ln 9}\,9^x + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.30     %-------------------------------------------------------------      \begin{align*}     \textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx &= \int n^{\frac{1 - n}{n}}\,x^{\frac{1 - n}{n}}\,dx     = n^{\frac{1 - n}{n}} \int x^{\frac{1 - n}{n}}\,dx = n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - n}{n} + 1}}{\frac{1 - n}{n} + 1} \right) + C \\[2mm]     &= n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - \cancel{n} + \cancel{n}}{n}}}{\frac{1 - \cancel{n} + \cancel{n}}{n}} \right) + C = n^{\frac{1 - n}{n}}\,n\,x^{1/n} + C = n^{1/n}\,x^{1/n} + C \\[2mm]     &= (nx)^{1/n} + C = \fboxsep=5pt\colorbox{gris!40}{$\sqrt[n]{nx} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.31     %-------------------------------------------------------------       \begin{align*}     \textbf{1.31)} \int 3^xe^x\,dx &= \int (3e)^x\,dx = \frac{(3e)^x}{\ln(3e)} + C = \frac{(3e)^x}{\ln 3 + \ln e} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3^xe^x}{\ln 3 + 1} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.32     %-------------------------------------------------------------       \begin{align*}     \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx &= \int {(a^{2/3})}^3 - 3{(a^{2/3})}^2x^{2/3} + 3a^{2/3}{(x^{2/3})}^2 - {(x^{2/3})}^3\,dx \\[3mm]     &= \int a^2 - 3a^{4/3}x^{2/3} + 3a^{2/3}x^{4/3} - x^2\,dx \\[3mm]     &= a^2 \int dx - 3a^{4/3} \int x^{2/3}\,dx + 3a^{2/3} \int x^{4/3}\,dx - \int x^2\,dx \\[3mm]     &= a^2x - 3a^{4/3}\left(\frac{x^{5/3}}{5/3} \right) + 3a^{2/3} \left(\frac{x^{7/3}}{7/3} \right) - \frac{x^3}{3} + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle a^2x - \frac{9a^{4/3}}{5}\,x^{5/3} + \frac{9a^{2/3}}{7}\,x^{7/3} - \frac{x^3}{3} + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.33     %-------------------------------------------------------------      \begin{align*}     \textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx &= \int \frac{x^{2m} - 2x^mx^n + x^{2n}}{\sqrt{x}}\,dx  \\[3mm]     &= \int x^{2m - 1/2}\,dx - 2 \int x^{m + n - 1/2}\,dx + \int x^{2n - 1/2}\,dx \\[3mm]     &= \frac{x^{2m - 1/2 + 1}}{2m - 1/2 + 1} - 2 \left(\frac{x^{m + n - 1/2 + 1}}{m + n - 1/2 + 1} \right) + \frac{x^{2n - 1/2 + 1}}{2n - 1/2 + 1} + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2 \left( \frac{x^{\frac{4m + 1}{2}}}{4m + 1} \right) - 4 \left( \frac{x^{\frac{2m + 2n + 1}{2}}}{2m + 2n + 1} \right) + 2 \left( \frac{x^{\frac{4n + 1}{2}}}{4n + 1} \right) + C$}     \end{align*}      %-------------------------------------------------------------     %                          EJERCICIO 1.34     %-------------------------------------------------------------      \begin{align*}     \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx &= \int \frac{a^{2x} - 2a^xb^x + b^{2x}}{a^xb^x}\,dx = \int \frac{a^{2x}}{a^xb^x}\,dx -2 \int \frac{\cancel{a^xb^x}}{\cancel{a^xb^x}}\,dx + \int \frac{b^{2x}}{a^xb^x}\,dx \\[2mm]     &= \int \left(\frac{a}{b} \right)^xdx - 2 \int dx + \int \left(\frac{b}{a} \right)^xdx \\[2mm]     &= \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln (a/b)} - 2x + \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln (b/a)} + C = \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln a - \ln b} - 2x - \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln a - \ln b} + C \\[2mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{\ln a - \ln b} \left[ \left(\frac{a}{b} \right)^x - \left(\frac{b}{a} \right)^x \right] - 2x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.35     %-------------------------------------------------------------       \begin{align*}     \textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy &= 4 \int y^3\,dy + 2 \int \frac{dy}{y^3} = \cancel{4}\left(\frac{y^4}{\cancel{4}} \right) - \cancel{2}\left(\frac{1}{- \cancel{2}y^2} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle y^4 + \frac{1}{y^2} + C$}     \end{align*}     \vspace{-0.65cm}       %-------------------------------------------------------------     %                          EJERCICIO 1.36     %-------------------------------------------------------------       \begin{align*}     \textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx &= \int \left(\frac{1}{\sqrt{2}\sen x} \right)^2 - 2\left(\frac{1}{\sqrt{2} \sen x} \right) + 1\,dx \\[3mm]     &= \int \frac{dx}{2 \sen^2x} - 2 \int \frac{dx}{\sqrt{2} \sen x} + \int dx = \frac{1}{2} \int \csc^2x\,dx - \sqrt{2} \int \csc x\,dx  + \int dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{2} \cot x - \sqrt{2} \ln \big|\csc x - \cot x \big| + x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.37     %-------------------------------------------------------------       \begin{align*}     \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta(1 + \tan^2\theta)}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta\,\cancel{\sec^2\theta}}{\cancel{\sec^2\theta}}\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \cos \theta + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.38     %-------------------------------------------------------------       \begin{align*}     \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx &= \int \frac{\sen x}{\cos^2x}\,dx = \int \frac{1}{\cos x}\,\frac{\sen x}{\cos x}\,dx = \int \sec \theta\,\tan \theta\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \sec \theta + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.39     %-------------------------------------------------------------       \begin{align*}     \textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx &= \int \frac{2\,\cancel{\sen x}\,\cos x}{\cancel{\sen x}}\,dx = \fboxsep=5pt\colorbox{gris!40}{$2\,\sen x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.40     %-------------------------------------------------------------       \begin{align*}     \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} &= \int \frac{dx}{(a - b) \left[\displaystyle \frac{a + b}{a - b} - x^2 \right]} = \frac{1}{a - b} \int \frac{dx}{\left(\sqrt{\frac{a + b}{a - b}} \right)^2 - x^2} \\[3mm]     &= \frac{1}{2(a - b)\displaystyle \sqrt{\frac{a + b}{a - b}}} \ln \left|\displaystyle \frac{\displaystyle \sqrt{\frac{a + b}{a - b}} + x}{\displaystyle \sqrt{\frac{a + b}{a - b}} - x} \right| + C \\[3mm]     &= \frac{1}{2\sqrt{(a + b)(a - b)}} \ln \left| \frac{\displaystyle  \frac{\sqrt{a + b} \ + \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}}{\displaystyle  \frac{\sqrt{a + b} \ - \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}} \right| + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{2 \sqrt{a^2 - b^2}} \ln \left| \frac{\sqrt{a + b} + \sqrt{a - b}x}{\sqrt{a + b} - \sqrt{a - b}x} \right| + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.41     %-------------------------------------------------------------        \begin{align*}     \textbf{1.41)} \int \frac{dx}{3x^2 + 5} &= \int \frac{dx}{3\left(x^2 + \frac{5}{3} \right)}     = \frac{1}{3} \int \frac{dx}{x^2 + \left(\sqrt{\frac{5}{3}} \right)^2} = \frac{1}{3} \left[\frac{1}{\sqrt{\frac{5}{3}}} \arctan \left(\frac{x}{\sqrt{\frac{5}{3}}} \right) \right] + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{\sqrt{3}}{3\sqrt{5}} \arctan \left(\frac{\sqrt{3} x}{\sqrt{5}} \right) + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.42     %-------------------------------------------------------------       \begin{align*}     \textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx &= \int {(\sqrt{x})}^2 - 2 \left(\cancel{\sqrt{x}} \frac{1}{\cancel{\sqrt{x}}} \right) + \left(\frac{1}{\sqrt{x}} \right)^2 dx = \int x^2\,dx - 2 \int dx + \int \frac{dx}{x} \\[3mm]     & = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{x^3}{3} - 2x + \ln \big| x \big| + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.43     %-------------------------------------------------------------       \begin{align*}     \textbf{1.43)}\int \left(y^2 - \frac{1}{y^2} \right)^3 dy &= \int {(y^2)}^3 - 3{(y^2)}^2\left(\frac{1}{y^2} \right) + 3y^2 \left(\frac{1}{y^2} \right)^2 - \left(\frac{1}{y^2} \right)^{\!\! 3} dy \\[3mm]     &= \int y^6\,dy - 3 \int y^2\,dy + 3 \int y^{-2}\,dy - \int y^{-6}\,dy \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{y^7}{7} - y^3 - \frac{3}{y} + \frac{1}{5y^5} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.44     %-------------------------------------------------------------       \begin{align*}     \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx &= \int \left( e^{x/a} \right)^2 - 2 e^{x/a} e^{- x/a} + \left( e^{- x/a} \right)^2 dx \\[3mm]     &= \int e^{2x/a} dx - 2 \int e^0\,dx + \int e^{-2x/a} dx \\[3mm]     &= \frac{e^{2x/a}}{\frac{2}{a}} - 2x - \frac{e^{- 2x/a}}{\frac{2}{a}} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{a}{2} \left[e^{2x/a} - e^{- 2x/a} \right] - 2x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.45     %-------------------------------------------------------------       \begin{align*}     \textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx &= \frac{\sqrt{5}}{5} \int x^{1/2}\,dx + \sqrt{5} \int x^{-1/2}\,dx = \frac{\sqrt{5}}{5} \left( \frac{2}{3} x^{3/2} \right) + \sqrt{5}(2 \sqrt{x}) + C \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2\sqrt{5}}{15}\,x^{3/2} + 2 \sqrt{5x} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.46     %-------------------------------------------------------------       \begin{align*}     \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt = 4 \int {(e^t)}^{-1/2} = - 4 \, \frac{e^{-t/2}}{1/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{8}{\sqrt{e^t}} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.47     %-------------------------------------------------------------       %\vspace*{-1cm}     \begin{align*}     \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}} &= \int \frac{dx}{\sqrt{5\left(\frac{7}{5} - x^2 \right)}} = \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(\sqrt{\frac{7}{5}} \right)^2 - x^2}} = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{\sqrt{5}}{5} \arcsen \left(\frac{\sqrt{5}x}{\sqrt{7}} \right) + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.48     %-------------------------------------------------------------       \begin{align*}     \textbf{1.48)} \int \frac{dx}{\sen x \cos x} &= \int \frac{\sen^2x + \cos^2x}{\sen x \cos x}\,dx = \int \frac{\sen^2x}{\sen x \cos x}\,dx + \int \frac{\cos^2x}{\sen x \cos x}\,dx = \int \tan x\,dx + \\[3mm]     & \int \cot x\,dx = - \ln \big| \cos x \big| + \ln \big| \sen x \big| + C = \ln \left| \frac{\sen x}{\cos x} \right| + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \ln \big| \tan x \big | + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.49     %-------------------------------------------------------------       \begin{align*}     \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta = \int \frac{d \theta}{\cos^2\theta} + \int \frac{\cancel{\cos^2 \theta}}{\cancel{\cos^2 \theta}}\,d \theta = \int \sec^2 \theta\,d \theta + \int d \theta     = \fboxsep=5pt\colorbox{gris!40}{$\tan \theta + \theta + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.50     %-------------------------------------------------------------       \begin{align*}     \textbf{1.50)} \int (\tan x + \sec x)^2\,dx &= \int \tan^2x + 2 \tan x\,\sec x + \sec^2x\,dx     \\[3mm]     &= \int \sec^2x - 1 + 2 \tan x\,\sec x + \sec^2x\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2\tan x + 2\sec x - x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.51     %-------------------------------------------------------------       \begin{align*}     \textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx = \int \left(x^{3/2} \right)^{1/2}dx = \int x^{3/4}dx = \frac{x^{7/4}}{7/4} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{4}{7}\,x^{7/4} + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.52     %-------------------------------------------------------------       \begin{align*}     \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx &= \int \frac{(3x^3 - 2) \cancel{(3x^3 + 2)}}{\cancel{(3x^3 + 2)}}\,dx = \int 3x^3\,dx - \int 2\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{4}\,x^4 - 2x + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.53     %-------------------------------------------------------------       \begin{align*}     \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi &= \int \frac{\displaystyle \frac{1}{\sen \phi}}{\displaystyle \frac{1}{\sen \phi} - \sen \phi}\,d\phi = \int \frac{\displaystyle \frac{1}{\cancel{\sen \phi}}}{\displaystyle \frac{1 - \sen^2\phi}{\cancel{\sen \phi}}}\,d \phi = \int \frac{1}{\cos^2\phi}\,d\phi = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \tan \phi + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.54     %-------------------------------------------------------------       \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Recuerde que la expresión $\int \frac{\sen x}{\cos^2x}\,dx$ ya había aparecido antes en el ejercicio 1.38 por lo que se omitieron algunos detalles de solución.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}     \vspace*{-0.25cm}     \begin{align*}     \textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx &= \int \frac{\sen x}{\cos x}\,dx + \int \frac{\tan x}{\cos x}\,dx = \int \tan x\,dx + \int \frac{\sen x}{\cos^2x}\,dx \\[3mm]     &= \int \tan x\,dx + \int \sec x \tan x\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \ln \big| \cos x \big| + \sec x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.55     %-------------------------------------------------------------      \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont La integral del ejercicio 1.56 es de la forma $a^x$ pero $a = 1$ por lo que no se puede aplicar la forma de la tabla, pero si las propiedades de los logaritmos.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}       \begin{align*}     \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx &= \frac{\sqrt{x(x - 1)} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx \\[3mm]     &= \int \frac{\sqrt{x}\sqrt{x - 1} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx = \int \frac{\cancel{\sqrt{x - 1}}(\sqrt{x} - e^x)}{\cancel{\sqrt{x - 1}}}\,dx \\[3mm]     &= \int \sqrt{x}\,dx - \int e^x\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{3}\,x^{3/2} - e^x + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.56     %-------------------------------------------------------------       \begin{align*}     \textbf{1.56)} \int 1^x\,dx = \int e^{\ln 1^x}dx = \int e^{x\,\ln 1}dx = \int e^0dx = \int dx = \fboxsep=5pt\colorbox{gris!40}{$x + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.57     %-------------------------------------------------------------       \begin{align*}     \textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx &= \int x^{5/3}x^{-4/3}(x^3 - 1)\,dx = \int x^{1/3}(x^3 - 1)\,dx  \\[3mm]     &= \int x^{10/3}dx - \int x^{1/3}dx = \frac{x^{13/3}}{13/3} + \frac{x^{4/3}}{4/3} + C     \end{align*}     \begin{align*}     = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{13}\,x^{13/3} + \frac{3}{4}\,x^{4/3} + C$}     \end{align*}        %-------------------------------------------------------------     %                          EJERCICIO 1.58     %-------------------------------------------------------------       \begin{align*}     \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx &= \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{(\sqrt{2x} - \sqrt{x})(\sqrt{2x} + \sqrt{x})}\,dx = \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{2x - x}\,dx     \\[3mm]     &= \int \frac{\sqrt{2}x^{3/2} + x^{3/2} - \sqrt{2}x^{1/2} - x^{1/2}}{x}\,dx \\[3mm]     &= \sqrt{2} \int x^{1/2}\,dx + \int x^{1/2}\,dx - \sqrt{2} \int x^{-1/2}\,dx - \int x^{- 1/2}\,dx \\[3mm]     &= ( \sqrt{2} + 1) \int x^{1/2}\,dx - (\sqrt{2} + 1) \int x^{-1/2}\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle (\sqrt{2} + 1) \left[\frac{2}{3}\,x^{3/2} - 2x^{1/2} \right] + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.59     %-------------------------------------------------------------       \begin{align*}     \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx &= \frac{\sqrt{5}}{\sqrt[3]{3}} \int \frac{x^{3/2}}{x^{1/3}}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \int x^{7/6}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \left( \frac{x^{13/6}}{13/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt{5}}{13 \sqrt[3]{3}}\,x^{13/6} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.60     %-------------------------------------------------------------       \begin{align*}     \textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx &= \int \frac{\tan x}{3 \sen x}\,dx - \int \frac{\sen^2x}{3 \sen x}\,dx + \int \frac{4 \cos x}{3 \sen x}\,dx \\[3mm]     &= \frac{1}{3} \int \frac{\cancel{\sen x}}{\cos x \, \cancel{\sen x}}\,dx - \frac{1}{3} \int \sen x\,dx + \frac{4}{3} \int \cot x\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{3}\,\ln \big| \sec x + \tan x \big| + \frac{1}{3} \cos x + \frac{4}{3} \ln \big| \sen x \big| + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.61     %-------------------------------------------------------------       \begin{align*}     \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx &= \int \left(\frac{1}{x} \right)^3 - 3 \left(\frac{1}{x} \right)^2 x + 3 \left(\frac{1}{x} \right)\,x^2 - x^3\,dx \\[3mm]     &= \int x^{-3}\,dx - 3 \int \frac{dx}{x} + 3 \int x\,dx - \int x^3\,dx \\[3mm]     &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{2\,x^2} - 3\,\ln \big| x \big| + \frac{3}{2}\,x^2 - \frac{x^4}{4} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.62     %-------------------------------------------------------------       \begin{align*}     \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx = \int \frac{\cancel{e^x}\,e^2}{\cancel{e^x}\,e}\,dx = \int e\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle ex + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.63     %-------------------------------------------------------------       \begin{align*}     \textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx = \int 8x^3 - 6x^2 - x^{-3}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2x^4 - 2x^3 + \frac{1}{2x^2} + C$}     \end{align*}       %-------------------------------------------------------------     %                          EJERCICIO 1.64     %-------------------------------------------------------------       \begin{align*}     \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx = \int \frac{4 \, \cancel{\ln x}}{\cancel{\ln x}}\,dx = 4 \int dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 4x + C$}     \end{align*}     \marginnote{     \colorbox{yellow!30}{     \begin{minipage}{2.5cm}     \begin{spacing}{0.55}     \fontsize{7}{14}\selectfont Observe con cuidado que en el ejercicio 1.64, la función logaritmo del numerador \emph{no está elevada a la cuarta potencia}, solo su argumento, por eso fue posible la aplicación de la propiedad y la posterior la simplificación de los términos.     \vspace{-8pt}     \end{spacing}     \end{minipage}}}       %-------------------------------------------------------------     %                          EJERCICIO 1.65     %-------------------------------------------------------------       \vspace*{-0.8cm}     \begin{align*}     \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx &= \int \frac{1 - \sen x}{(1 + \sen x)(1 - \sen x)}\,dx = \int \frac{1 - \sen x}{1 - \sen^2x}\,dx     = \int \frac{1 - \sen x}{\cos^2x}\,dx     \\[3mm]     &= \int \sec^2x\,dx - \int \frac{\sen x}{\cos^2x}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \tan x - \sec x + C$}     \end{align*}     \end{document}

Stefan Kottwitz
You forgot to add leqno as option to \documentclass, at the beginning.