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\documentclass[a4paper,12pt]{article}
\usepackage[top=.2in, bottom=.5in,left=1in,right=1in]{geometry}
\usepackage{amsmath}
\usepackage{cancel}
\begin{document}
\begin{align*}
~& \quad \; \, \int \frac{x+1}{x^{3}+1}\,dx\\\\
~&= \int \frac{x+1}{x(x^{2}+1)}\,dx \\\\
~&= \int \left ( \frac{A}{x}+\frac{Bx+C}{x^{2}+1} \right )\,dx \\\\
x+1 &= A(x^{2}+1)+(Bx+C)(x)\\
x+1 &= Ax^{2}+A+Bx^{2}+Cx\\
x+1 &= Ax^{2}+Bx^{2}+A+Cx\\
0x^{2}+x+1 &= x^{2}(A+B)+Cx+A\\
~& A+B{\,=\,}0, \quad C=1, \quad A=1\\
~& \;1+B{\,=\,}0\\
~& \qquad B{\,=\,}-1\\\\
~&= \int \frac{1}{x}+\frac{-x+1}{x^{2}+1}\,dx\\\\
~&= \int \frac{1}{x}\,dx+\int \frac{-x+1}{x^{2}+1}\,dx\\\\
~&= \int \frac{1}{x}\,dx-\int \frac{x+1}{x^{2}+1}\,dx\\\\
~&= \int \frac{1}{x}\,dx-\int \frac{x}{x^{2}+1}+\frac{1}{x^{2}+1}\,dx\\\\
~&= \int \frac{1}{x}\,dx-\int \frac{x}{x^{2}+1}\,dx+\int\frac{1}{x^{2}+1}\,dx\\\\
~&= ln\left | x \right |-\int \frac{x}{x^{2}+1}\,dx+\int\frac{1}{x^{2}+1}\,dx\\\\
\text{for }\int\frac{x}{x^{2}+1}\,dx & \quad \text{let } w=x^{2}+1,\quad \frac{dw}{dx}{\,=\,}2x,\quad \frac{dw}{2x}{\,=\,}dx \\\\
~&= ln\left | x \right |-\int \frac{\cancel{x}}{x}\frac{dw}{2\cancel{x}}\,+\int\frac{1}{x^{2}+1}\,dx\\\\
~&= ln\left | x \right | - \frac{1}{2}\int\frac{1}{x}\,dx+\int \frac{1}{x^{2}+1}\,dx\\\\
~&= ln\left | x \right | - \frac{1}{2}\,ln\left | x \right |+\int \frac{1}{x^{2}+1}\,dx\\\\
\end{align*}
\begin{align*}
\text{for }\int\frac{1}{x^{2}+1}\,dx\; & \text{ Use the table }\int\frac{1}{x^{2}+a^{2}}\,dx{\;=\;}\frac{1}{a}\,arctan \biggl ( \frac{x}{a} \biggr )\,+\,C\\\\
~&= ln\left | x \right | - \frac{1}{2}\,ln\left | x \right |+\int \frac{1}{x^{2}+1}\,dx\\\\
~&= ln\left | x \right | - \frac{1}{2}\,ln\left | x \right |+arctan(x)\,+\,C\\\\
\end{align*}
\end{document}
