Yes, I did it, I tested your code and followed your instructions but now, my code:
Code: Select all
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\ifoot*{Ing. Daniel A. Veliz V.}
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\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]
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\\[-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]
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Variable de integración \end{varwidth}};
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Constante de integración \end{varwidth}};
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\\[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}
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{\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
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{\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}
