100 Integrals - Rewritten

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{bigints}
\DeclareMathOperator{\arcsec}{arcsec}
\DeclareMathOperator{\arccot}{arccot}
\DeclareMathOperator{\arccsc}{arccsc}
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\title{100 Integrals}
\author{Josiah Rondaris}
\begin{document}
\maketitle
\section{Basic Integrals to Remember}
\begin{align*}
&\int x^ndx=\frac{1}{n+1}x^{n+1}+C \\
&\int\sin\left(x\right)dx=-\cos\left(x\right)+C \\
&\int\cos\left(x\right)dx=\sin\left(x\right)+C \\
&\int\tan\left(x\right)dx=-\ln\left|\cos\left(x\right)\right|+C\mbox{ or }\ln\left|\sec\left(x\right)\right|+C \\
&\int\csc\left(x\right)dx=-\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+C \\
&\int\sec\left(x\right)dx=\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|+C \\
&\int\cot\left(x\right)dx=\ln\left|\sin\left(x\right)\right|+C\mbox{ or }-\ln\left|\csc\left(x\right)\right|+C \\
&\int a^xdx=\frac{1}{\ln\left(a\right)}a^x+C \\
&\int e^xdx=e^x+C \\
&\int\log_a\left(x\right)dx=x\log_a\left(x\right)-\frac{1}{\ln\left(a\right)}x+C \\
&\int\ln\left(x\right)dx=x\ln\left(x\right)-x+C
\end{align*}
\section{Advanced Integrals to Remember}
\begin{align*}
&\int\sec^3(x)dx=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C \\
&\int\arcsin(x)dx=x\arcsin(x)+\sqrt{1-x^2}+C \\
&\int\arccos(x)dx=x\arccos(x)-\sqrt{1-x^2}+C \\
&\int\arctan(x)dx=x\arctan(x)-\frac{1}{2}\ln(x^2+1)+C \\
&\int\sqrt{x^2+a^2}dx=\frac{1}{2}x\sqrt{x^2+a^2}+\frac{a^2}{2}\ln\left|\sqrt{x^2+a^2}+x\right|+C \\
&\int\sqrt{x^2-a^2}dx=\frac{1}{2}x\sqrt{x^2-a^2}-\frac{a^2}{2}\ln\left|\sqrt{x^2-a^2}+x\right|+C \\
&\int\sqrt{a^2-x^2}dx=\frac{a^2}{2}\arcsin\left(\frac{x}{a}\right)+\frac{1}{2}x\sqrt{a^2-x^2}+C \\
&\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C \\
&\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C \\
&\int\frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C \\
&\int\frac{1}{\sqrt{x^2+a^2}}dx=\ln\left|x+\sqrt{x^2+a^2}\right|+C \\
&\int\frac{1}{\sqrt{x^2-a^2}}dx=\ln\left|x+\sqrt{x^2-a^2}\right|+C \\
&\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin\left(\frac{x}{a}\right)+C \\
&\int\sin^n(x)dx=-\frac{1}{n}\sin^{n-1}(x)\cos(x)+\frac{n-1}{n}\int\sin^{n-2}(x)dx \\
&\int\cos^n(x)dx=\frac{1}{n}\cos^{n-1}(x)\sin(x)+\frac{n-1}{n}\int\cos^{n-2}(x)dx
\end{align*}
\newpage
\section{The 100 Integrals}
\subsection*{Problem 1}
$$\int\frac{1}{\sqrt{x}\left(1+x\right)}dx$$
Substituting
\begin{align*}
u&=\sqrt{x} \\
du&=\frac{1}{2\sqrt{x}}dx
\end{align*}
yields:
$$=2\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=2\arctan(u)+C$$
Undoing the substitution(s):
$$=2\arctan\left(\sqrt{x}\right)+C$$
\subsection*{Problem 2}
$$\int\frac{\sec^2\left(x\right)}{1+\tan\left(x\right)}dx$$
Substituting
\begin{align*}
u&=1+\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int\frac{1}{u}du$$
This integral is standard:
$$=\ln|u|+C$$
Undoing the substitution(s):
$$=\ln\left|1+\tan(x)\right|+C$$
\subsection*{Problem 3}
$$\int\sin\left(x\right)\sec\left(x\right)dx$$
Rewriting the integral:
$$=\int\frac{\sin(x)}{\cos(x)}dx$$
Substituting
\begin{align*}
u&=\cos(x) \\
du&=-\sin(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\int\ln|u|+C$$
Undoing the substitution(s):
$$=-\ln\left|\cos(x)\right|+C$$
\subsection*{Problem 4}
$$\int\frac{\csc\left(x\right)\cot\left(x\right)}{1+\csc^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\csc(x) \\
du&=-\csc(x)\cot(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=-\arctan(u)+C$$
Undoing the substitution(s):
$$=-\arctan(\csc(x))+C$$
\subsection*{Problem 5}
$$\int\frac{\tan\left(x\right)}{\cos^2\left(x\right)}dx$$
Rewriting the integral:
$$=\int\tan(x)\sec^2(x)dx$$
Substituting
\begin{align*}
u&=\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
Yields:
$$=\int udu$$
This integral is standard:
$$=\frac{1}{2}u^2+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\tan^2(x)+C$$
\subsection*{Problem 6}
$$\int\csc^4\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int\csc^2(x)\cdot\csc^2(x)dx \\
&=\int\csc^2(x)(\cot^2(x)+1)dx
\end{align*} 
Substituting
\begin{align*}
u&=\cot(x) \\
du&=-\csc^2(x)dx
\end{align*}
Yields:
\begin{align*}
&=-\int(u^2+1)du \\
&=-\int u^2du-\int du
\end{align*}
Both integrals are standard:
$$=-\frac{1}{3}u^3-u+C$$
Undoing the substitution(s):
$$=-\frac{1}{3}\cot^3(x)-\cot(x)+C$$
\subsection*{Problem 7}
$$\int x\tan^2\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int x(\sec^2(x)-1)dx \\
&=\int x\sec^2(x)dx- \int xdx
\end{align*}
Using integration by parts where
\begin{align*}
u&=x &v&=\tan(x) \\
du&=dx &dv&=\sec^2(x)dx
\end{align*}
yields:
$$=x\tan(x)-\int\tan(x)dx-\int xdx$$
Both integrals are standard:
$$=x\tan(x)+\ln|\cos(x)|-\frac{1}{2}x^2+C$$
\subsection*{Problem 8}
$$\int x^2\cos^2\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int x^2\left(\frac{1+\cos(2x)}{2}\right)dx \\
&=\frac{1}{2}\int x^2(1+\cos(2x))dx \\
&=\frac{1}{2}\int x^2dx+\frac{1}{2}\int x^2\cos(2x)dx
\end{align*}
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 Sign & D & I \\
 \hline
 $+$ & $x^2$ & $\cos(2x)$ \\ 
 $-$ & $2x$ & $\frac{1}{2}\sin(2x)$ \\
 $+$ & $2$ & $-\frac{1}{4}\cos(2x)$ \\
 $-$ & $0$ & $-\frac{1}{8}\sin(2x)$ \\
 \hline
\end{tabular}
\end{center}
yields:
\begin{align*}
&=\frac{1}{6}x^3+\frac{1}{2}\left(\frac{1}{2}x^2\sin(2x)+\frac{1}{2}x\cos(2x)-\frac{1}{4}\sin(2x)\right)+C \\
&=\frac{1}{6}x^3+\frac{1}{4}x^2\sin(2x)+\frac{1}{4}x\cos(2x)-\frac{1}{8}\sin(2x)+C
\end{align*}
\subsection*{Problem 9}
$$\int x^5\sqrt{2-x^3}dx$$
Rewriting the integral:
$$=\int x^2\cdot x^3\sqrt{2-x^3}dx$$
Substituting
\begin{align*}
u&=2-x^3 \\
2-u&=x^3 \\
-du&=3x^2dx
\end{align*}
yields:
\begin{align*}
&=-\frac{1}{3}\int(2-u)\sqrt{u}du \\
&=-\frac{1}{3}\int\left(2\sqrt{u}-u^{3/2}\right)du \\
&=-\frac{2}{3}\int\sqrt{u}du+\frac{1}{3}\int u^{3/2}du
\end{align*}
Both integrals are standard:
$$=-\frac{4}{9}u^{3/2}+\frac{2}{15}u^{5/2}+C$$
Undoing the substitution(s):
$$=-\frac{4}{9}(2-x^3)^{3/2}+\frac{2}{15}(2-x^3)^{5/2}+C$$
\subsection*{Problem 10}
$$\int\frac{1}{\sqrt{x^2+4}}dx$$
This integral is standard:
$$=\ln\left|x+\sqrt{x^2+4}\right|+C$$
\subsection*{Problem 11}
$$\int\frac{x^2}{\sqrt{25+x^2}}dx$$
Rewriting the integral:
\begin{align*}
=\int\frac{x^2}{\sqrt{25\big(1+\left(\frac{x}{5}\right)^2\big)}}dx \\
=\frac{1}{5}\int\frac{x^2}{\sqrt{1+(\frac{x}{5})^2}}dx
\end{align*}
Substituting
\begin{align*}
\tan\theta&=\frac{x}{5} \\
5\tan\theta&=x \\
5\sec^2\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{(5\tan\theta)^2}{\sqrt{1+\tan^2\theta}}\sec^2\theta d\theta \\
&=25\int\frac{\tan^2\theta\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta \\
&=25\int(\sec^2\theta-1)\sec\theta d\theta \\
&=25\int(\sec^3\theta-\sec\theta)d\theta \\
&=25\int\sec^3\theta d\theta-25\int\sec\theta d\theta
\end{align*}
Both integrals are standard:
\begin{align*}
&=25\left(\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|\right)-25\ln|\sec\theta+\tan\theta|+C \\
&=\frac{25}{2}\sec\theta\tan\theta+\frac{25}{2}\ln|\sec\theta+\tan\theta|-25\ln|\sec\theta+\tan\theta|+C \\
&=\frac{25}{2}\sec\theta\tan\theta-\frac{25}{2}\ln|\sec\theta+\tan\theta|+C
\end{align*}
Undoing the substitution(s):
\begin{align*}
&=\frac{25}{2}\cdot\frac{\sqrt{x^2+25}}{5}\cdot\frac{x}{5}-\frac{25}{2}\ln\left|\frac{\sqrt{x^2+25}}{5}+\frac{x}{5}\right|+C \\
&=\frac{1}{2}x\sqrt{x^2+25}-\frac{25}{2}\ln\left|\sqrt{x^2+25}+x\right|+C
\end{align*}
\subsection*{Problem 12}
$$\int\cos\left(x\right)\sqrt{4-\sin^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\sqrt{4-u^2}du$$
This integral is standard:
$$=2\arcsin\left(\frac{u}{2}\right)+\frac{1}{2}u\sqrt{4-u^2}+C$$
Undoing the substitution(s):
$$=2\arcsin\left(\frac{\sin(x)}{2}\right)+\frac{1}{2}\sin(x)\sqrt{4-\sin^2(x)}+C$$
\subsection*{Problem 13}
$$\int\frac{1}{x^2-x+1}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{1}{x^2-x+\frac{1}{4}+\frac{3}{4}}dx \\
&=4\int\frac{1}{4x^2-4x+1+3}dx \\
&=4\int\frac{1}{(2x-1)^2+3}
\end{align*}
Substituting
\begin{align*}
u&=2x-1 \\
du&=2dx
\end{align*}
yields:
$$=2\int\frac{1}{u^2+3}du$$
This integral is standard:
$$=\frac{2}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right)+C$$
Undoing the substitution(s):
$$=\frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C$$
\subsection*{Problem 14}
$$\int\sqrt{x^2+x+1}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{x^2+x+\frac{1}{4}+\frac{3}{4}}dx \\
&=\frac{1}{2}\int\sqrt{4x^2+4x+1+3}dx \\
&=\frac{1}{2}\int\sqrt{(2x+1)^2+3}dx
\end{align*}
Substituting:
\begin{align*}
u&=2x+1 \\
du&=2dx
\end{align*}
yields:
$$=\frac{1}{4}\int\sqrt{u^2+3}du$$
This integral is standard:
\begin{align*}
&=\frac{1}{4}\left(\frac{1}{2}u\sqrt{u^2+3}+\frac{3}{2}\ln\left|\sqrt{u^2+3}+u\right|\right)+C \\
&=\frac{1}{8}u\sqrt{u^2+3}+\frac{3}{8}\ln\left|\sqrt{u^2+3}+u\right|+C
\end{align*}
Undoing the substitution(s):
$$=\frac{2x+1}{8}\sqrt{4x^2+4x+4}+\frac{3}{8}\ln\left|\sqrt{4x^2+4x+4}+2x+1\right|+C$$
\subsection*{Problem 15}
$$\int\frac{5x+31}{3x^2-4x+11}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{5x-\frac{10}{3}+\frac{103}{3}}{3x^2-4x+11}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+\frac{103}{9}\int\frac{1}{x^2-\frac{4}{3}x+\frac{11}{3}}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+\frac{103}{9}\int\frac{1}{x^2-\frac{4}{3}x+\frac{4}{9}+\frac{29}{9}}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+103\int\frac{1}{9x^2-12x+4+29}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+103\int\frac{1}{(3x-2)^2+29}
\end{align*}
Substituting:
\begin{align*}
u&=3x^2-4x+11 &v&=3x-2 \\
du&=(6x-4)dx &dv&=3dx
\end{align*}
yields:
$$=\frac{5}{6}\int\frac{1}{u}+\frac{103}{3}\int\frac{1}{v^2+29}dv$$
Both integrals are standard:
$$=\frac{5}{6}\ln|u|+\frac{103}{3\sqrt{29}}\arctan\left(\frac{v}{\sqrt{29}}\right)+C$$
Undoing the substitution(s):
$$=\frac{5}{6}\ln\left|3x^2-4x+11\right|+\frac{103}{3\sqrt{29}}\arctan\left(\frac{3x-2}{\sqrt{29}}\right)+C$$
\subsection*{Problem 16}
$$\int\frac{x^4+1}{x^2+2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^4+2x^2-2x^2-4+5}{x^2+2}dx \\
&=\int\frac{x^2(x^2+2)-2(x^2+2)+5}{x^2+2}dx \\
&=\int\left(x^2-2+\frac{5}{x^2+2}\right)dx \\
&=\int x^2dx-2\int dx+5\int\frac{1}{x^2+2}dx
\end{align*}
All three integrals are standard:
$$=\frac{1}{3}x^3-2x+\frac{5}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)+C$$
\subsection*{Problem 17}
$$\int\frac{1}{5+4\cos\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
du&=\frac{1+u^2}{2}dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=2\int\frac{1}{5+4\left(\frac{1-u^2}{1+u^2}\right)}\cdot\frac{1}{1+u^2}du \\
&=2\int\frac{1}{5(1+u^2)+4(1-u^2)}du \\
&=2\int\frac{1}{5+5u^2+4-4u^2}du \\
&=2\int\frac{1}{9+u^2}du
\end{align*}
This integral is standard:
$$=\frac{2}{3}\arctan\left(\frac{u}{3}\right)+C$$
Undoing the substitution(s):
$$=\frac{2}{3}\arctan\left(\frac{\tan(\frac{x}{2})}{3}\right)+C$$
\subsection*{Problem 18}
$$\int\frac{\sqrt{x}}{1+x}dx$$
Substituting
\begin{align*}
\tan\theta&=\sqrt{x} \\
\tan^2\theta&=x \\
2\tan\theta\sec^2\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=2\int\frac{\tan\theta}{1+\tan^2\theta}\tan\theta\sec^2\theta d\theta \\
&=2\int\frac{\tan^2\theta\sec^2\theta}{\sec^2\theta}d\theta \\
&=2\int\left(\sec^2\theta-1\right)d\theta \\
&=2\int\sec^2\theta-2\int d\theta
\end{align*}
Both integrals are standard:
$$=2\tan\theta-2\theta+C$$
Undoing the substitution(s):
$$=2\sqrt{x}-2\arctan\left(\sqrt{x}\right)+C$$
\subsection*{Problem 19}
$$\int\frac{\cos\left(x\right)}{4-\sin^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\frac{1}{4-u^2}du$$
This integral is standard:
$$=\frac{1}{4}\ln\left|\frac{2+u}{2-u}\right|+C$$
Undoing the substitution(s):
$$=\frac{1}{4}\ln\left|\frac{2+\sin(x)}{2-\sin(x)}\right|+C$$
\subsection*{Problem 20}
$$\int\frac{\cos\left(2x\right)}{\cos\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{2\cos^2(x)-1}{\cos(x)}dx \\
&=\int(2\cos(x)-\sec(x))dx \\
&=2\int\cos(x)dx-\int\sec(x)dx
\end{align*}
Both integrals are standard:
$$=2\sin(x)-\ln|\sec(x)+\tan(x)|+C$$
\subsection*{Problem 21}
$$\int\frac{\tan\left(x\right)}{\ln\left(\cos\left(x\right)\right)}dx$$
Substituting
\begin{align*}
u&=-\ln(\cos(x)) \\
du&=\tan(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\ln|u|+C$$
Undoing the substitution(s):
$$=-\ln\left|-\ln(\cos(x))\right|+C$$
\subsection*{Problem 22}
$$\int\frac{x^7}{\sqrt{1-x^4}}dx$$
Rewriting the integral:
$$=\int\frac{x^3\cdot x^4}{\sqrt{1-x^4}}dx$$
Substituting:
\begin{align*}
u&=1-x^4 \\
1-u&=x^4 \\
-du&=4x^3dx
\end{align*}
Yields:
\begin{align*}
&=-\frac{1}{4}\int\frac{1-u}{\sqrt{u}}du \\
&=-\frac{1}{4}\int\left(u^{-1/2}-\sqrt{u}\right)du \\
&=-\frac{1}{4}\int u^{-1/2}du+\frac{1}{4}\int\sqrt{u}du
\end{align*}
Both integrals are standard:
$$=-\frac{1}{2}\sqrt{u}+\frac{1}{6}u^{3/2}+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}\sqrt{1-x^4}+\frac{1}{6}(1-x^4)^{3/2}+C$$
\subsection*{Problem 23}
$$\int\ln\left(1+x\right)dx$$
Substituting
\begin{align*}
u&=1+x \\
du&=dx 
\end{align*}
Yields:
$$=\int\ln(u)du$$
This integral is standard:
$$=u\ln(u)-u+C$$
Undoing the substitution(s):
$$=(1+x)\ln(1+x)-(1+x)+C$$
\subsection*{Problem 24}
$$\int x\arcsec\left(x\right)dx$$
Using integration by parts where
\begin{align*}
u&=\arcsec(x) &v&=\frac{1}{2}x^2 \\
du&=\frac{1}{x\sqrt{x^2-1}}dx &dv&=xdx
\end{align*}
yields:
$$=\frac{1}{2}x^2\arcsec(x)-\frac{1}{2}\int\frac{x}{\sqrt{x^2-1}}dx$$
Substituting
\begin{align*}
w&=x^2-1 \\
dw&=2xdx
\end{align*}
yields:
$$=-\frac{1}{4}\int\frac{1}{\sqrt{w}}dw$$
This integral is standard:
$$=\frac{1}{2}x^2\arcsec(x)-\frac{1}{2}\sqrt{w}+C$$
Undoing the substitution(s):
$$=\frac{1}{2}x^2\arcsec(x)-\frac{1}{2}\sqrt{x^2-1}+C$$
\subsection*{Problem 25}
$$\int\sqrt{x^2+9}dx$$
This integral is standard:
$$=\frac{1}{2}x\sqrt{x^2+9}+\frac{9}{2}\ln\left|\sqrt{x^2+9}+x\right|+C$$
\subsection*{Problem 26}
$$\int\frac{x^2}{\sqrt{4-x^2}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^2}{\sqrt{4(1-\frac{x^2}{4})}}dx \\
&=\frac{1}{2}\int\frac{x^2}{\sqrt{1-(\frac{x}{2})^2}}dx
\end{align*}
Substituting
\begin{align*}
\sin\theta&=\frac{x}{2} \\
2\sin\theta&=x \\
2\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{(2\sin\theta)^2\cos\theta}{\sqrt{1-\sin^2\theta}}d\theta \\
&=4\int\frac{\sin^2\theta\cos\theta}{\sqrt{\cos^2\theta}}d\theta \\
&=4\int\sin^2\theta d\theta \\
&=4\int\left(\frac{1-\cos2\theta}{2}\right)d\theta \\
&=2\int(1-\cos2\theta)d\theta \\
&=2\int d\theta-2\int\cos2\theta d\theta
\end{align*}
Substituting
\begin{align*}
u&=2\theta \\
du&=2d\theta 
\end{align*}
yields:
$$=2\int d\theta-\int\cos(u)du$$
Both integrals are standard:
$$=2\theta-\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=2\theta-\sin(2\theta)+C \\
&=2\theta-2\sin\theta\cos\theta+C \\
&=2\arcsin\left(\frac{x}{2}\right)-2\cdot\frac{x}{2}\cdot\frac{\sqrt{4-x^2}}{2}+C \\
&=2\arcsin\left(\frac{x}{2}\right)-\frac{1}{2}x\sqrt{4-x^2}+C
\end{align*}
\subsection*{Problem 27}
$$\int\sqrt{2x-x^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{1-1+2x-x^2}dx \\
&=\int\sqrt{1-(1-2x+x^2)}dx \\
&=\int\sqrt{1-(1-x)^2}dx
\end{align*}
Substituting:
\begin{align*}
u&=1-x \\
du&=-dx
\end{align*}
yields:
$$=-\int\sqrt{1-u^2}du$$
This integral is standard:
$$=-\frac{1}{2}\arcsin(u)-\frac{1}{2}u\sqrt{1-u^2}+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}\arcsin(1-x)-\frac{1}{2}(1-x)\sqrt{2x-x^2}+C$$
\subsection*{Problem 28}
$$\int\frac{4x-2}{x^3-x}dx$$
Factoring the integral:
$$=\int\frac{4x-2}{x(x-1)(x+1)}dx$$
Using partial fraction decomposition
$$=\int\left(\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\right)dx$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{2}{x}+\frac{1}{x-1}+\frac{-3}{x+1}\right)dx \\
&=2\int\frac{1}{x}dx+\int\frac{1}{x-1}dx-3\int\frac{1}{x+1}dx
\end{align*}
Substituting
\begin{align*}
u&=x-1 &v&=x+1 \\
du&=dx &dv&=dx
\end{align*}
yields:
$$2\int\frac{1}{x}dx+\int\frac{1}{u}du-3\int\frac{1}{v}dv$$
All integrals are standard:
\begin{align*}
&=2\ln|x|+\ln|u|-3\ln|v|+C \\
&=\ln\left|\frac{x^2u}{v^3}\right|+C
\end{align*}
Undoing the substitution(s):
$$=\ln\left|\frac{x^2(x-1)}{(x+1)^3}\right|+C$$
\subsection*{Problem 29}
$$\int\frac{x^4}{x^2-2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^4-2x^2+2x^2-4+4}{x^2-2}dx \\
&=\int\left(\frac{x^4-2x^2}{x^2-2}+\frac{2x^2-4}{x^2-2}+\frac{4}{x^2-2}\right)dx \\
&=\int\frac{x^2(x^2-2)}{x^2-2}dx+\int\frac{2(x^2-2)}{x^2-2}dx+4\int\frac{1}{x^2-2}dx \\
&=\int x^2dx+2\int dx+4\int\frac{1}{x^2-2}dx
\end{align*}
All three integrals are standard:
$$=\frac{1}{3}x^3+2x+\sqrt{2}\ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C$$
\subsection*{Problem 30}
$$\int\frac{\sec\left(x\right)\tan\left(x\right)}{\sec\left(x\right)+\sec^2\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\tan(x)}{1+\sec(x)}\cdot\frac{\cos(x)}{\cos(x)}dx \\
&=\int\frac{\sin(x)}{\cos(x)+1}dx
\end{align*}
Substituting
\begin{align*}
u&=\cos(x)+1 \\
du&=-\sin(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\ln\left|u\right|+C$$
Undoing the substitution(s):
$$=-\ln\left|\cos(x)+1\right|+C$$
\subsection*{Problem 31}
$$\int\frac{x}{\left(x^2+2x+2\right)^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x}{(x^2+2x+1+1)^2}dx \\
&=\int\frac{x}{((x+1)^2+1)^2}dx
\end{align*}
Substituting
\begin{align*}
\tan\theta&=x+1 \\
\tan\theta-1&=x \\
\sec^2\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{\tan\theta-1}{(\tan^2\theta+1)^2}\sec^2\theta d\theta \\
&=\int\frac{\tan\theta-1}{\sec^2\theta}\cdot\frac{\cos^2\theta}{\cos^2\theta}d\theta \\
&=\int\left(\sin\theta\cos\theta-\cos^2\theta\right) \\
&=\int\left(\sin\theta\cos\theta-\frac{1+\cos2\theta}{2}\right)d\theta \\
&=\int\sin\theta\cos\theta d\theta-\frac{1}{2}\int d\theta-\frac{1}{2}\int\cos2\theta d\theta
\end{align*}
Substituting:
\begin{align*}
u&=\sin\theta &v&=2\theta \\
du&=\cos\theta d\theta &dv&=2d\theta
\end{align*}
yields:
$$=\int udu-\frac{1}{2}\int d\theta-\frac{1}{4}\int\cos(u)du$$
All three integrals are standard:
$$=\frac{1}{2}u^2-\frac{1}{2}\theta-\frac{1}{4}\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{2}\sin^2\theta-\frac{1}{2}\theta-\frac{1}{4}\sin2\theta+C \\
&=\frac{1}{2}\sin^2\theta-\frac{1}{2}\theta-\frac{1}{2}\sin\theta\cos\theta+C \\
&=\frac{1}{2}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right)^2-\frac{1}{2}\arctan(x+1)-\frac{1}{2}\cdot\frac{x+1}{\sqrt{x^2+2x+2}}\cdot\frac{1}{\sqrt{x^2+2x+2}}+C \\
&=\frac{x^2+2x+1}{2(x^2+2x+2)}-\frac{x+1}{2(x^2+2x+2)}-\frac{1}{2}\arctan(x+1)+C \\
&=\frac{x^2+x}{2x^2+4x+4}-\frac{1}{2}\arctan(x+1)+C
\end{align*}
\subsection*{Problem 32}
$$\int\frac{x^{1/3}}{x^{1/2}+x^{1/4}}dx$$
Substituting
\begin{align*}
u^3&=\sqrt[4]{x} \\
u^4&=\sqrt[3]{x} \\
u^6&=\sqrt{x} \\
u^{12}&=x\\
12u^{11}du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{u^4}{u^6+u^3}12u^{11}du \\
&=12\int\frac{u^{12}}{u^3+1}du
\end{align*}
Performing long division on the integral yields:
\begin{align*}
&=12\int\left(u^9-u^6+u^3-1+\frac{1}{u^3+1}\right)du \\
&=12\int u^9du-12\int u^6du+12\int u^3du-12\int du+12\int\frac{1}{u^3+1}du
\end{align*}
Factoring the last integral:
$$=12\int\frac{1}{(u+1)(u^2-u+1)}du$$
Using partial fraction decomposition
$$=12\int\left(\frac{A}{u+1}+\frac{Bu+C}{u^2-u+1}\right)du$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=12\int\left(\frac{\frac{1}{3}}{u+1}+\frac{-\frac{1}{3}u+\frac{2}{3}}{u^2-u+1}\right)du \\
&=4\int\frac{1}{u+1}du-4\int\frac{u-2}{u^2-u+1}du
\end{align*}
Rewriting the second integral:
\begin{align*}
&=4\int\frac{1}{u+1}du-4\int\frac{u-\frac{1}{2}-\frac{3}{2}}{u^2-u+1}du \\
&=4\int\frac{1}{u+1}du-4\int\frac{u-\frac{1}{2}}{u^2-u+1}du+4\int\frac{\frac{3}{2}}{u^2-u+1}du \\
&=4\int\frac{1}{u+1}du-2\int\frac{2u-1}{u^2-u+1}du+6\int\frac{1}{u^2-u+\frac{1}{4}+\frac{3}{4}}du \\
&=4\int\frac{1}{u+1}du-2\int\frac{2u-1}{u^2-u+1}du+24\int\frac{1}{4u^2-4u+1+3}du \\
&=4\int\frac{1}{u+1}du-2\int\frac{2u-1}{u^2-u+1}du+24\int\frac{1}{(2u-1)^2+3}du
\end{align*}
Substituting
\begin{align*}
v&=u+1 &w&=u^2-u+1 &y&=2u-1 \\
dv&=du &dw&=(2u-1)du &dy&=2du
\end{align*}
yields:
$$=4\int\frac{1}{v}dv-2\int\frac{1}{w}dw+12\int\frac{1}{y^2+3}dy$$
All three integrals are standard:
\begin{align*}
&=4\ln|v|-2\ln|w|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C \\
&=\ln\left|v^4\right|-\ln\left|w^2\right|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C \\
&=\ln\left|\frac{v^4}{w^2}\right|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C
\end{align*}
The previous four integrals are also standard:
$$=\frac{6}{5}u^{10}-\frac{12}{7}u^7+3u^4-12u+\ln\left|\frac{v^4}{w^2}\right|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{6}{5}u^{10}-\frac{12}{7}u^7+3u^4-12u+\ln\left|\frac{(u+1)^4}{(u^2-u+1)^2}\right|+4\sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right)+C \\
&=\frac{6}{5}x^{5/6}-\frac{12}{7}x^{7/12}+3x^{1/3}-12x^{1/12}+\cdots \\
&+\ln\left|\frac{(x^{1/12}+1)^4}{(x^{1/6}-x^{1/12}+1)^2}\right|+4\sqrt{3}\arctan\left(\frac{2x^{1/12}-1}{\sqrt{3}}\right)+C
\end{align*}
\subsection*{Problem 33}
$$\int\frac{1}{1+\cos\left(2x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{1}{1+2\cos^2(x)-1}dx \\
&=\frac{1}{2}\int\sec^2(x)dx
\end{align*}
This integral is standard:
$$=\frac{1}{2}\tan(x)+C$$
\subsection*{Problem 34}
$$\int\frac{\sec\left(x\right)}{\tan\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\sec(x)}{\tan(x)}\cdot\frac{\cos(x)}{\cos(x)}dx \\
&=\int\frac{1}{\sin(x)}dx \\
&=\int\csc(x)dx
\end{align*}
This integral is standard:
$$=-\ln\left|\cot(x)+\csc(x)\right|+C$$
\subsection*{Problem 35}
$$\int\sec^3\left(x\right)\tan^3\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int\sec^2(x)\tan^2(x)\sec(x)\tan(x)dx \\
&=\int\sec^2(x)(\sec^2(x)-1)\sec(x)\tan(x) \\
&=\int(\sec^4(x)-\sec^2(x))\sec(x)\tan(x)
\end{align*}
Substituting
\begin{align*}
u&=\sec(x) \\
du&=\sec(x)\tan(x)dx
\end{align*}
yields:
\begin{align*}
&=\int\left(u^4-u^2\right)du \\
&=\int u^4du-\int u^2du
\end{align*}
Both integrals are standard:
$$=\frac{1}{5}u^5-\frac{1}{3}u^3+C$$
Undoing the substitution(s):
$$=\frac{1}{5}\sec^5(x)-\frac{1}{3}\sec^3(x)+C$$
\subsection*{Problem 36}
$$\int x^2\arctan\left(x\right)dx$$
Using integration by parts where
\begin{align*}
u&=\arctan(x) &v&=\frac{1}{3}x^3 \\
du&=\frac{1}{x^2+1}dx &dv&=x^2dx
\end{align*}
yields:
$$=\frac{1}{3}x^3\arctan(x)-\frac{1}{3}\int\frac{x^3}{x^2+1}dx$$
\subsection*{Problem 37}
$$\int x\ln^3\left(x\right)dx$$
Substituting
\begin{align*}
u&=\ln(x) \\
e^u&=x \\
e^udu&=dx
\end{align*}
yields:
$$=\int u^3e^{2u}du$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 Sign & D & I \\
 \hline
 $+$ & $u^3$ & $e^{2u}$ \\ 
 $-$ & $3u^2$ & $\frac{1}{2}e^{2u}$ \\
 $+$ & $6u$ & $\frac{1}{4}e^{2u}$ \\
 $-$ & $6$ & $\frac{1}{8}e^{2u}$ \\
 $+$ & $0$ & $\frac{1}{16}e^{2u}$ \\
 \hline
\end{tabular}
\end{center}
yields:
$$=\frac{1}{2}u^3e^{2u}-\frac{3}{4}u^2e^{2u}+\frac{3}{4}ue^{2u}-\frac{3}{8}e^{2u}+C$$
Undoing the substitution(s):
$$=\frac{1}{2}x^2\ln^3(x)-\frac{3}{4}x^2\ln^2(x)+\frac{3}{4}x^2\ln(x)-\frac{3}{8}x^2+C$$
\subsection*{Problem 38}
$$\int\frac{1}{x\sqrt{1+x^2}}dx$$
Substituting
\begin{align*}
\tan\theta&=x \\
\sec^2\theta d\theta&dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{\sec^2\theta}{\tan\theta\sqrt{1+\tan^2\theta}}d\theta \\
&=\int\frac{\sec^2\theta}{\tan\theta\sqrt{\sec^2\theta}}d\theta \\
&=\int\frac{\sec\theta}{\tan\theta}d\theta \\
&=\int\frac{\sec\theta}{\tan\theta}\cdot\frac{\cos\theta}{\cos\theta}d\theta \\
&=\int\frac{1}{\sin\theta}d\theta \\
&=\int\csc\theta d\theta
\end{align*}
This integral is standard:
$$=-\ln\left|\cot\theta+\csc\theta\right|+C$$
Undoing the substitution(s):
\begin{align*}
&=-\ln\left|\frac{1}{x}+\frac{\sqrt{x^2+1}}{x}\right|+C \\
&=-\ln\left|\frac{1+\sqrt{x^2+1}}{x}\right|+C \\
&=\ln\left|\frac{x}{1+\sqrt{x^2+1}}\right|+C
\end{align*}
\subsection*{Problem 39}
$$\int e^x\sqrt{1+e^{2x}}dx$$
Substituting
\begin{align*}
\tan\theta&=e^x \\
\sec^2\theta d\theta&=e^xdx
\end{align*}
yields:
\begin{align*}
&=\int\sec^2\theta\sqrt{1+\tan^2\theta}d\theta \\
&=\int\sec^2\theta\sqrt{\sec^2\theta}d\theta \\
&=\int\sec^3\theta d\theta
\end{align*}
This integral is standard:
$$=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{2}e^x\sqrt{1+e^{2x}}+\frac{1}{2}\ln\left|\sqrt{1+e^{2x}}+e^x\right|+C
\end{align*}
\subsection*{Problem 40}
$$\int\frac{x}{4x-x^2}dx$$
Rewriting the integral:
$$=\int\frac{1}{4-x}dx$$
Substituting
\begin{align*}
u&=4-x \\
du&=-dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\ln\left|u\right|+C$$
Undoing the substitution(s):
$$=-\ln\left|4-x\right|+C$$
\subsection*{Problem 41}
$$\int\frac{1}{x^3\sqrt{x^2-9}}dx$$
Rewriting the integral:
\begin{align*}
=&\int\frac{1}{x^3\sqrt{9\left(\frac{x^2}{9}-1\right)}}dx \\
=&\frac{1}{3}\int\frac{1}{x^3\sqrt{\left(\frac{x}{3}\right)^2-1}}dx
\end{align*}
Substituting
\begin{align*}
\sec\theta&=\frac{x}{3} \\
3\sec\theta&=x \\
3\sec\theta\tan\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{3}\int\frac{3\sec\theta\tan\theta}{(3\sec\theta)^3\sqrt{\sec^2\theta-1}}d\theta \\
&=\frac{1}{27}\int\frac{\tan\theta}{\sec^2\theta\sqrt{\tan^2\theta}}d\theta \\
&=\frac{1}{27}\int\cos^2\theta d\theta \\
&=\frac{1}{54}\int\left(1+\cos2\theta\right)d\theta \\
&=\frac{1}{54}\int d\theta+\frac{1}{54}\int\cos2\theta d\theta
\end{align*}
Substituting
\begin{align*}
u&=2\theta \\
du&2d\theta
\end{align*}
yields:
$$=\frac{1}{54}\int d\theta+\frac{1}{108}\int\cos(u)du$$
Both integrals are standard:
$$=\frac{1}{54}\theta+\frac{1}{108}\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{54}\theta+\frac{1}{108}\sin2\theta+C \\
&=\frac{1}{54}\theta+\frac{1}{54}\sin\theta\cos\theta+C \\
&=\frac{1}{54}\arcsec\left(\frac{x}{3}\right)+\frac{1}{54}\cdot\frac{\sqrt{x^2-9}}{x}\cdot\frac{3}{x}+C \\
&=\frac{1}{54}\arcsec\left(\frac{x}{3}\right)+\frac{\sqrt{x^2-9}}{18x^2}+C
\end{align*}
\subsection*{Problem 42}
$$\int\frac{x}{\left(7x+1\right)^{17}}dx$$
Substituting
\begin{align*}
u&=7x+1 \\
\frac{1}{7}u-\frac{1}{7}&=x \\
\frac{1}{7}du&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{7}\int\frac{\frac{1}{7}u-\frac{1}{7}}{u^{17}}du \\
&=\frac{1}{49}\int\frac{u-1}{u^{17}}du \\
&=\frac{1}{49}\int\left(u^{-16}-u^{-17}\right)du \\
&=\frac{1}{49}\int u^{-16}du-\frac{1}{49}\int u^{-17}du
\end{align*}
Both integrals are standard:
$$=-\frac{1}{735u^{15}}+\frac{1}{784u^{16}}+C$$
Undoing the substitution(s):
$$=-\frac{1}{735(7x+1)^{15}}+\frac{1}{784(7x+1)^{16}}+C$$
\subsection*{Problem 43}
$$\int\frac{4x^2+x+1}{4x^3+x}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{4x^2+1+x}{x(4x^2+1)}dx \\
&=\int\left(\frac{4x^2+1}{x(4x^2+1)}+\frac{x}{x(4x^2+1)}\right)dx \\
&=\int\left(\frac{1}{x}+\frac{1}{4x^2+1}\right)dx \\
&=\int\frac{1}{x}dx+\int\frac{1}{4x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=2x \\
du&=2dx
\end{align*}
yields:
$$=\int\frac{1}{x}dx+\frac{1}{2}\int\frac{1}{u^2+1}du$$
Both integrals are standard:
$$=\ln|x|+\frac{1}{2}\arctan(u)+C$$
Undoing the substitution(s):
$$=\ln\left|x\right|+\frac{1}{2}\arctan(2x)+C$$
\subsection*{Problem 44}
$$\int\frac{4x^3-x+1}{x^3+1}dx$$
Performing long division on the integral yields:
\begin{align*}
&=\int\left(4-\frac{x+3}{x^3+1}\right)dx \\
&=4\int dx-\int\frac{x+3}{(x+1)(x^2-x+1)}dx
\end{align*}
Using partial fraction decomposition
$$=-\int\left(\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\right)$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=-\int\left(\frac{\frac{2}{3}}{x+1}+\frac{-\frac{2}{3}x+\frac{7}{3}}{x^2-x+1}\right)dx
\\
&=-\frac{1}{3}\int\left(\frac{2}{x+1}-\frac{2x-7}{x^2-x+1}\right)dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-7}{x^2-x+1}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1-6}{x^2-x+1}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\left(\frac{2x-1}{x^2-x+1}-\frac{6}{x^2-x+1}\right)dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx-2\int\frac{1}{x^2-x+\frac{1}{4}+\frac{3}{4}}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx-8\int\frac{1}{4x^2-4x+1+3}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx-8\int\frac{1}{(2x-1)^2+3}dx
\end{align*}
Substituting
\begin{align*}
u&=x+1 &v&=x^2-x+1 &w&=2x-1 \\
du&=dx &dv&=(2x-1)dx &dw&=2dx
\end{align*}
yields:
$$=-\frac{2}{3}\int\frac{1}{u}du+\frac{1}{3}\int\frac{1}{v}dv-4\int\frac{1}{w^2+3}dw$$
All three integrals are standard:
\begin{align*}
&=-\frac{2}{3}\ln|u|+\frac{1}{3}\ln|v|-\frac{4}{\sqrt{3}}\arctan\left(\frac{w}{\sqrt{3}}\right)+C \\
&=-\frac{1}{3}\ln\left|u^2\right|+\frac{1}{3}\ln|v|-\frac{4}{\sqrt{3}}\arctan\left(\frac{w}{\sqrt{3}}\right)+C \\
&=\frac{1}{3}\ln\left|\frac{v}{u^2}\right|-\frac{4}{\sqrt{3}}\arctan\left(\frac{w}{\sqrt{3}}\right)+C
\end{align*}
Undoing the substitution(s):
$$=4x+\frac{1}{3}\ln\left|\frac{x^2-x+1}{(x+1)^2}\right|-\frac{4}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C$$
\subsection*{Problem 45}
$$\int\tan^2\left(x\right)\sec\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int(\sec^2(x)-1)\sec(x)dx \\
&=\int(\sec^3(x)-\sec(x))dx \\
&=\int\sec^3(x)dx-\int\sec(x)dx
\end{align*}
Both integrals are standard:
\begin{align*}
&=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln|\sec(x)+\tan(x)-\ln|\sec(x)+\tan(x)|+C \\
&=\frac{1}{2}\sec(x)\tan(x)-\frac{1}{2}\ln|\sec(x)-\tan(x)|+C
\end{align*}
\subsection*{Problem 46}
$$\int\frac{x^2+2x+2}{\left(x+1\right)^3}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^2+2x+1+1}{(x+1)^3}dx \\
&=\int\left(\frac{x^2+2x+1}{(x+1)^3}+\frac{1}{(x+1)^3}\right)dx \\
&=\int\left(\frac{(x+1)^2}{(x+1)^3}+\frac{1}{(x+1)^3}\right)dx \\
&=\int\frac{1}{x+1}dx+\int\frac{1}{(x+1)^3}dx
\end{align*}
Substituting
\begin{align*}
u&=x+1 \\
du&=dx
\end{align*}
Yields:
$$=\int\frac{1}{u}du+\int\frac{1}{u^3}du$$
Both integrals are standard:
$$=\ln|u|-\frac{1}{2u^2}+C$$
Undoing the substitution(s):
$$=\ln|x+1|-\frac{1}{2(x+1)^2}+C$$
\subsection*{Problem 47}
$$\int\frac{x^4+2x+2}{x^5+x^4}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^4+2(x+1)}{x^5+x^4}dx \\
&=\int\left(\frac{x^4}{x^4(x+1)}+\frac{2(x+1)}{x^4(x+1)}\right)dx \\
&=\int\frac{1}{x+1}dx+2\int\frac{1}{x^4}dx
\end{align*}
Substituting
\begin{align*}
u&=x+1 \\
du&=dx
\end{align*}
yields:
$$=\int\frac{1}{u}du+2\int\frac{1}{x^4}dx$$
Both integrals are standard:
$$=\ln|u|-\frac{2}{3x^3}+C$$
Undoing the substitution(s):
$$=\ln|x+1|-\frac{2}{3x^3}+C$$
\subsection*{Problem 48}
$$\int\frac{8x^2-4x+7}{\left(x^2+1\right)\left(4x+1\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{8x^2+8-4x-1}{(x^2+1)(4x+1)}dx \\
&=\int\frac{8(x^2+1)-(4x+1)}{(x^2+1)(4x+1)} \\
&=\int\left(\frac{8(x^2+1)}{(x^2+1)(4x+1)}-\frac{4x+1}{(x^2+1)(4x+1)}\right)dx \\
&=\int\left(\frac{8}{4x+1}-\frac{1}{x^2+1}\right)dx \\
&=8\int\frac{1}{4x+1}-\int\frac{1}{x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=4x+1 \\
du&=4dx
\end{align*}
yields:
$$=2\int\frac{1}{u}du-\int\frac{1}{x^2+1}dx$$
Both integrals are standard:
$$=2\ln|u|-\arctan(x)+C$$
Undoing the substitution(s):
$$=2\ln|4x+1|-\arctan(x)+C$$
\subsection*{Problem 49}
$$\int\frac{3x^5-x^4+2x^3-12x^2-2x+1}{\left(x^3-1\right)^2}dx$$
Factoring the integral:
\begin{align*}
&=\int\frac{3x^5-x^4+2x^3-12x^2-2x+1}{((x-1)(x^2+x+1))^2}dx \\
&=\int\frac{3x^5-x^4+2x^3-12x^2-2x+1}{(x-1)^2(x^2+x+1)^2}dx
\end{align*}
Using partial fraction decomposition
$$=\int\left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+x+1}+\frac{Ex+F}{(x^2+x+1)^2}\right)dx$$
where $A,B,C,D,E,F\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{1}{x-1}-\frac{1}{(x-1)^2}+\frac{2x+1}{x^2+x+1}+\frac{4x+2}{(x^2+x+1)^2}\right)dx \\
&=\int\frac{1}{x-1}dx-\int\frac{1}{(x-1)^2}dx+\int\frac{2x+1}{x^2+x+1}dx+2\int\frac{2x+1}{(x^2+x+1)^2}dx
\end{align*}
Substituting
\begin{align*}
u&=x-1 &v&=x^2+x+1 \\
du&=dx &dv&=(2x+1)dx
\end{align*}
yields:
$$=\int\frac{1}{u}du-\int\frac{1}{u^2}du+\int\frac{1}{v}dv+2\int\frac{1}{v^2}dv$$
All four integrals are standard:
\begin{align*}
&=\ln|u|+\frac{1}{u}+\ln|v|-\frac{2}{v}+C \\
&=\ln|uv|+\frac{v}{uv}-\frac{2u}{uv}+C \\
&=\ln|uv|+\frac{v-2u}{uv}+C
\end{align*}
Undoing the substitution(s):
\begin{align*}
&=\ln|(x-1)(x^2+x+1)|+\frac{x^2+x+1-2(x-1)}{(x-1)(x^2+x+1)}+C \\
&=\ln\left|x^3-1\right|+\frac{x^2-x+3}{x^3-1}+C
\end{align*}
\subsection*{Problem 50}
$$\int\frac{x}{x^4+4x^2+8}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x}{x^4+4x^2+4+4}dx \\
&=\int\frac{x}{(x^2+2)^2+4}dx
\end{align*}
Substituting
\begin{align*}
u&=x^2+2 \\
du&=2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{u^2+4}du$$
This integral is standard:
$$=\frac{1}{4}\arctan\left(\frac{u}{2}\right)+C$$
Undoing the substitution(s):
$$=\frac{1}{4}\arctan\left(\frac{x^2+2}{2}\right)+C$$
\subsection*{Problem 51}
$$\int\frac{1}{4+5\cos\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1}{4+5\left(\frac{1-u^2}{1+u^2}\right)}\cdot\frac{2}{1+u^2}du \\
&=2\int\frac{1}{4(1+u^2)+5(1-u^2)}du \\
&=2\int\frac{1}{4+4u^2+5-5u^2}du \\
&=2\int\frac{1}{9-u^2}du
\end{align*}
This integral is standard:
$$=\frac{1}{3}\ln\left|\frac{3+u}{3-u}\right|+C$$
Undoing the substitution(s):
$$=\frac{1}{3}\ln\left|\frac{3+\tan(\frac{x}{2})}{3-\tan(\frac{x}{2})}\right|+C$$
\subsection*{Problem 52}
$$\int\frac{\left(1+x^{2/3}\right)^{3/2}}{x^{1/3}}dx$$
Substituting
\begin{align*}
u&=1+x^{2/3} \\
du&=\frac{2}{3}x^{-1/3}dx
\end{align*}
yields:
$$=\frac{3}{2}\int u^{3/2}du$$
This integral is standard:
$$=\frac{3}{5}u^{5/2}+C$$
Undoing the substitution(s):
$$=\frac{3}{5}(1+x^{2/3})^{5/2}+C$$
\subsection*{Problem 53}
$$\int\frac{\arcsin^2\left(x\right)}{\sqrt{1-x^2}}dx$$
Substituting
\begin{align*}
u&=\arcsin(x) \\
du&=\frac{1}{\sqrt{1-x^2}}dx
\end{align*}
yields:
$$=\int u^2du$$ \\
This integral is standard:
$$=\frac{1}{3}u^3+C$$
Undoing the substitution(s):
$$=\frac{1}{3}\arcsin^3(x)+C$$
\subsection*{Problem 54}
$$\int\frac{1}{x^{3/2}\left(1+x^{1/3}\right)}dx$$
Substituting
\begin{align*}
u&=x^{1/6} \\
u^2&=x^{1/3} \\
u^6&=x \\
u^9&=x^{3/2} \\
6u^5du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1}{u^9(1+u^2)}6u^5du \\
&=6\int\frac{1}{u^4(1+u^2)}du
\end{align*}
Using partial fraction decomposition
$$=\int\left(\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u^3}+\frac{D}{u^4}+\frac{Eu+F}{u^2+1}\right)du$$
where $A,B,C,D,E,F\in\mathbb{R}$ yields:
\begin{align*}
&=6\int\left(-\frac{1}{u^2}+\frac{1}{u^4}+\frac{1}{u^2+1}\right)du \\
&=-6\int\frac{1}{u^2}du+6\int\frac{1}{u^4}du+6\int\frac{1}{u^2+!}du
\end{align*}
All three integrals are standard:
$$=\frac{6}{u}-\frac{2}{u^3}+6\arctan(u)+C$$
Undoing the substitution(s):
$$=\frac{6}{x^{1/6}}-\frac{2}{\sqrt{x}}+6\arctan\left(x^{1/6}\right)+C$$
\subsection*{Problem 55}
$$\int\tan^3\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int\tan^2(x)\tan(x)dx \\
&=\int(\sec^2(x)-1)\tan(x)dx \\
&=\int(\sec^2(x)\tan(x)-\tan(x))dx \\
&=\int\sec^2(x)\tan(x)dx-\int\tan(x)dx
\end{align*}
Substituting
\begin{align*}
u&=\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int udu-\int\tan(x)dx$$
Both integrals are standard:
$$=\frac{1}{2}u^2+\ln|\cos(x)|+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$
\subsection*{Problem 56}
$$\int\sin^2\left(x\right)\cos^4\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int(1-\cos^2(x))\cos^4(x)dx \\
&=\int(\cos^4(x)-\cos^6(x))dx \\
&=\int\cos^4(x)dx-\int\cos^6(x)dx
\end{align*}
Using the reduction rule for $\cos^n(x)$, it's known that:
\begin{align*}
\int\cos^4(x)dx&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{4}\int\cos^2(x)dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{4}\int\frac{1+\cos(2x)}{2}dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\int(1+\cos(2x))dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\int dx+\frac{3}{8}\int\cos(2x)dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}x+\frac{3}{16}\sin(2x)+C \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\sin(x)\cos(x)+\frac{3}{8}x+C
\end{align*}
and
$$\int\cos^6(x)dx=\frac{1}{6}\cos^5(x)\sin(x)+\frac{5}{6}\int\cos^4(x)dx$$
Rewriting the integral:
\begin{align*}
&=\int\cos^4(x)dx-\frac{1}{6}\cos^5(x)\sin(x)-\frac{5}{6}\int\cos^4(x)dx \\
&=-\frac{1}{6}\cos^5(x)\sin(x)+\frac{1}{6}\int\cos^4(x)dx \\
&=-\frac{1}{6}\cos^5(x)\sin(x)+\frac{1}{6}\left(\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\sin(x)\cos(x)+\frac{3}{8}x+C\right) \\
&=-\frac{1}{6}\cos^5(x)\sin(x)+\frac{1}{24}\cos^3(x)\sin(x)+\frac{1}{16}\sin(x)\cos(x)+\frac{1}{16}x+C
\end{align*}
\subsection*{Problem 57}
$$\int\frac{xe^{x^2}}{1+e^{2x^2}}dx$$
Substituting
\begin{align*}
u&=e^{x^2} \\
du&=2xe^{x^2}
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=\frac{1}{2}\arctan(u)+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\arctan\left(e^{x^{2}}\right)+C$$
\subsection*{Problem 58}
$$\int\frac{\cos^3\left(x\right)}{\sqrt{\sin\left(x\right)}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\cos(x)\cos^2(x)}{\sqrt{\sin(x)}}dx \\
&=\int\frac{\cos(x)(1-\sin^2(x))}{\sqrt{\sin(x)}}dx
\end{align*}
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1-u^2}{\sqrt{u}}du \\
&=\int\left(\frac{1}{\sqrt{u}}-u^{3/2}\right)du \\
&=\int\frac{1}{\sqrt{u}}du-\int u^{3/2}du
\end{align*}
Both integrals are standard:
$$=2\sqrt{u}-\frac{2}{5}u^{5/2}+C$$
Undoing the substitution(s):
$$=2\sqrt{\sin(x)}-\frac{2}{5}\left(\sin(x)\right)^{5/2}+C$$
\subsection*{Problem 59}
$$\int x^3e^{-x^2}dx$$
Rewriting the integral:
$$=\int x\cdot x^2e^{-x^2}dx$$
Substituting
\begin{align*}
u&=-x^2 \\
-u&=x^2 \\
-du&=2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int ue^udu$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 Sign & D & I \\
 \hline
 $+$ & $u$ & $e^u$ \\ 
 $-$ & 1 & $e^u$ \\
 $+$ & 0 & $e^u$ \\
 \hline
\end{tabular}
\end{center}
yields:
$$=\frac{1}{2}ue^u-\frac{1}{2}e^u+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}x^2e^{-x^2}-\frac{1}{2}e^{-x^2}+C$$
\subsection*{Problem 60}
$$\int\sin\left(\sqrt{x}\right)dx$$
Substituting
\begin{align*}
u&=\sqrt{x} \\
u^2&=x \\
2udu&=dx
\end{align*}
yields:
$$=2\int u\sin(u)du$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 Sign & D & I \\
 \hline
 $+$ & $u$ & $\sin(u)$ \\ 
 $-$ & 1 & $-\cos(u)$ \\
 $+$ & 0 & $-\sin(u)$ \\
 \hline
\end{tabular}
\end{center}
yields:
$$=-2u\cos(u)+2\sin(u)+C$$
Undoing the substitution(s):
$$=-2\sqrt{x}\cos\left(\sqrt{x}\right)+2\sin\left(\sqrt{x}\right)+C$$
\subsection*{Problem 61}
$$\int\frac{\arcsin\left(x\right)}{x^2}dx$$
Substituting
\begin{align*}
\sin\theta&=x \\
\theta&=\sin(x) \\
\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{\theta\cos\theta}{\sin^2\theta}d\theta \\
&=\int u\cot\theta\csc\theta d\theta
\end{align*}
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 Sign & D & I \\
 \hline
 $+$ & $\theta$ & $\cot\theta\csc\theta$ \\ 
 $-$ & 1 & $-\csc\theta$ \\
 $+$ & 0 & $\ln|\csc\theta+\cot\theta|$ \\
 \hline
\end{tabular}
\end{center}
yields:
$$=-\theta\csc\theta-\ln\left|\csc\theta+\cot\theta\right|+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{\arcsin(x)}{x}-\ln\left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|+C \\
&=-\frac{\arcsin(x)}{x}-\ln\left|\frac{1+\sqrt{1-x^2}}{x}\right|+C
\end{align*}
\subsection*{Problem 62}
$$\int\sqrt{x^2-9}dx$$
This integral is standard:
$$=\frac{1}{2}x\sqrt{x^2-9}-\frac{9}{2}\ln\left|\sqrt{x^2-9}+x\right|+C$$
\subsection*{Problem 63}
$$\int x^2\sqrt{1-x^2}dx$$
Substituting
\begin{align*}
\sin\theta&=x \\
\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\sin^2\theta\cos\theta\sqrt{1-\sin^2\theta}d\theta \\
&=\int\sin^2\theta\cos\theta\sqrt{\cos^2\theta}d\theta \\
&=\int\sin^2\theta\cos^2\theta d\theta \\
&=\frac{1}{4}\int4\sin^2\theta\cos^2\theta d\theta \\
&=\frac{1}{4}\int(2\sin\theta\cos\theta)^2 d\theta \\
&=\frac{1}{4}\int\sin^22\theta d\theta \\
&=\frac{1}{4}\int\left(\frac{1-\cos4\theta}{2}\right)d\theta \\
&=\frac{1}{8}\int(1-\cos4\theta)d\theta \\
&=\frac{1}{8}\int d\theta-\frac{1}{8}\int\cos4\theta d\theta
\end{align*}
Substituting
\begin{align*}
u&=4\theta \\
du&=4d\theta
\end{align*}
yields:
$$=\frac{1}{8}\int d\theta-\frac{1}{32}\int\cos(u)du$$
Both integrals are standard:
$$=\frac{1}{8}\theta-\frac{1}{32}\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{8}\theta-\frac{1}{32}\sin4\theta+C \\
&=\frac{1}{8}\theta-\frac{1}{16}\sin2\theta\cos2\theta+C \\
&=\frac{1}{8}\theta-\frac{1}{8}\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)+C \\
&=\frac{1}{8}\arcsin(x)-\frac{1}{8}x\sqrt{1-x^2}\left(\left(\sqrt{1-x^2}\right)^2-x^2\right)+C \\
&=\frac{1}{8}\arcsin(x)-\frac{1}{8}x\sqrt{1-x^2}(1-x^2-x^2)+C \\
&=\frac{1}{8}\arcsin(x)-\frac{1}{8}(x-2x^3)\sqrt{1-x^2}+C
\end{align*}
\subsection*{Problem 64}
$$\int x\sqrt{2x-x^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\left(1-\frac{1}{2}(2-2x)\right)\sqrt{2x-x^2}dx \\
&=\int\sqrt{2x-x^2}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx \\
&=\int\sqrt{1-1+2x-x^2}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx \\
&=\int\sqrt{1-(1-2x+x^2)}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx \\
&=\int\sqrt{1-(1-x)^2}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx
\end{align*}
Substituting
\begin{align*}
u&=1-x &v&=2x-x^2 \\
du&=-dx &dv&=(2-2x)dx
\end{align*}
$$=-\int\sqrt{1-u^2}du-\frac{1}{2}\int\sqrt{v}dv$$
Both integrals are standard:
$$=-\frac{1}{2}\arcsin(u)-\frac{1}{2}u\sqrt{1-u^2}-\frac{1}{3}v^{3/2}+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}\arcsin(1-x)-\frac{1-x}{2}\sqrt{2x-x^2}-\frac{1}{3}(2x-x^2)^{3/2}+C$$
\subsection*{Problem 65}
$$\int\frac{x-2}{4x^2+4x+1}dx$$
Rewriting the integral:
$$=\int\frac{x-2}{(2x+1)^2}dx$$
Substituting
\begin{align*}
u&=2x+1 \\
\frac{u-1}{2}&=x \\
\frac{1}{2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{2}\int\frac{\frac{1}{2}u-\frac{5}{2}}{u^2}du \\
&=\frac{1}{4}\int\frac{u-5}{u^2}du \\
&=\frac{1}{4}\int\left(\frac{1}{u}-\frac{5}{u^2}\right)du \\
&=\frac{1}{4}\int\frac{1}{u}du-\frac{5}{4}\int\frac{1}{u^2}du
\end{align*}
Both integrals are standard:
$$=\frac{1}{4}\ln|u|+\frac{5}{4u}+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{4}\ln|2x+1|+\frac{5}{4(2x+1)}+C \\
&=\frac{1}{4}\ln|2x+1|+\frac{5}{8x+4}+C
\end{align*}
\subsection*{Problem 66}
$$\int\frac{2x^2-5x-1}{x^3-2x^2-x+2}dx$$
Factoring the integral:
\begin{align*}
&=\int\frac{2x^2-5x-1}{x^2(x-2)-(x-2)}dx \\
&=\int\frac{2x^2-5x-1}{(x^2-1)(x-2)}dx \\
&=\int\frac{2x^2-5x-1}{(x-1)(x+1)(x-2)}dx
\end{align*}
Using partial fraction decomposition
$$=\int\left(\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}\right)dx$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{2}{x-1}+\frac{1}{x+1}+\frac{-1}{x-2}\right)dx \\
&=2\int\frac{1}{x-1}dx+\int\frac{1}{x+1}dx-\int\frac{1}{x-2}dx
\end{align*}
Substituting
\begin{align*}
u_1&=x-1 &u_2&=x+1 &u_3&=x-2 \\
du_1&=dx &du_2&=dx &du_3&=dx
\end{align*}
yields:
$$=2\int\frac{1}{u_1}du_1+\int\frac{1}{u_2}du_2-\int\frac{1}{u_3}du_3$$
All three integrals are standard:
\begin{align*}
&=2\ln\left|u_1\right|+\ln\left|u_2\right|-\ln\left|u_3\right|+C \\
&=\ln\left|u_1^2\right|+\ln\left|u_2\right|-\ln\left|u_3\right|+C \\
&=\ln\left|\frac{u_1^2u_2}{u_3}\right|+C
\end{align*}
Undoing the substitution(s):
$$=\ln\left|\frac{(x-1)^2(x+1)}{x-2}\right|+C$$
\subsection*{Problem 67}
$$\int\frac{e^{2x}}{e^{2x}-1}dx$$
Substituting
\begin{align*}
u&=e^{2x}-1 \\
du&=2e^{2x}dx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{u}du$$
This integral is standard:
$$=\frac{1}{2}\ln|u|+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\ln\left|e^{2x}-1\right|+C$$
\subsection*{Problem 68}
$$\int\frac{\cos\left(x\right)}{\sin^2\left(x\right)-3\sin\left(x\right)+2}dx$$
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\frac{1}{u^2-3u+2}du$$
Factoring the integral:
$$=\int\frac{1}{(u-2)(u-1)}du$$
Using partial fraction decomposition
$$=\int\left(\frac{A}{u-2}+\frac{B}{u-1}\right)du$$
where $A,B\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{1}{u-2}+\frac{-1}{u-1}\right)du \\
&=\int\frac{1}{u-2}du-\int\frac{1}{u-1}du
\end{align*}
Substituting
\begin{align*}
v_1&=u-2 &v_2&=u-1 \\
dv_1&=du &dv_2&=du
\end{align*}
yields:
$$=\int\frac{1}{v_1}dv_1-\int\frac{1}{v_2}dv_2$$
Both integrals are standard:
\begin{align*}
&=\ln\left|v_1\right|-\ln\left|v_2\right|+C \\
&=\ln\left|\frac{v_1}{v_2}\right|+C
\end{align*}
Undoing the substitution(s):
$$=\ln\left|\frac{\sin(x)-2}{\sin(x)-1}\right|+C$$
\subsection*{Problem 69}
$$\int\frac{2x^3+3x^2+4}{\left(x+1\right)^4}dx$$
Substituting
\begin{align*}
u&=x+1 \\
u-1&=x \\
du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{2(u-1)^3+3(u-1)^2+4}{u^4}du \\
&=\int\frac{2(u^3-3u^2+3u-1)+3(u^2-2u+1)+4}{u^4}du \\
&=\int\frac{2u^3-6u^2+6u-2+3u^2-6u+3+4}{u^4}du \\
&=\int\frac{2u^3-3u^2+5}{u^4}du \\
&=\int\left(\frac{2}{u}-\frac{3}{u^2}+\frac{5}{u^4}\right)du \\
&=2\int\frac{1}{u}du-3\int\frac{1}{u^2}du+5\int\frac{1}{u^4}du
\end{align*}
All three are standard integrals:
$$=2\ln|u|+\frac{3}{u}-\frac{5}{3u^3}+C$$
Undoing the substitution(s):
$$=2\ln|x+1|+\frac{3}{x+1}-\frac{5}{3(x+1)^3}+C$$
\subsection*{Problem 70}
$$\int\frac{\sec^2\left(x\right)}{\tan^2\left(x\right)+2\tan\left(x\right)+2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\sec^2(x)}{\tan^2(x)+2\tan(x)+1+1}dx \\
&=\int\frac{\sec^2(x)}{(\tan(x)+1)^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=\tan(x)+1 \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int\frac{1}{u^2+1}du$$
This integral is standard:
$$=\arctan(u)+C$$
Undoing the substitution(s):
$$=\arctan(\tan(x)+1)+C$$
\subsection*{Problem 71}
$$\int\frac{x^3+x^2+2x+1}{x^4+2x^2+1}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^3+x+x+x^2+1}{(x^2+1)^2}dx \\
&=\int\frac{x+x(x^2+1)+x^2+1}{(x^2+1)^2}dx \\
&=\int\left(\frac{x}{(x^2+1)^2}+\frac{x(x^2+1)}{(x^2+1)^2}+\frac{x^2+1}{(x^2+1)^2}\right)dx \\
&=\int\frac{x}{(x^2+1)^2}dx+\int\frac{x}{x^2+1}dx+\int\frac{1}{x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=x^2+1 \\
du&=2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{u^2}du+\frac{1}{2}\int\frac{1}{u}du$$
Both are standard integrals:
$$=-\frac{1}{2u}+\frac{1}{2}\ln|u|+\arctan(x)+C$$
Undoing the substitution(s):
$$=-\frac{1}{2x^2+2}+\frac{1}{2}\ln\left|x^2+1\right|+\arctan(x)+C$$
\subsection*{Problem 72}
$$\int\frac{3+\cos\left(x\right)}{2-\cos\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=-\int\frac{\cos(x)+3}{\cos(x)-2}dx \\
&=-\int\frac{\cos(x)-2+5}{\cos(x)-2}dx \\
&=-\int\left(\frac{\cos(x)-2}{\cos(x)-2}+\frac{5}{\cos(x)-2}\right)dx \\
&=-\int dx-5\int\frac{1}{\cos(x)-2}dx
\end{align*}
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=-5\int\frac{1}{\frac{1-u^2}{1+u^2}-2}\cdot\frac{2}{1+u^2}du \\
&=-10\int\frac{1}{1-u^2-2(1+u^2)}du \\
&=-10\int\frac{1}{1-u^2-2-2u^2}du \\
&=-10\int\frac{1}{-1-3u^2}du \\
&=10\int\frac{1}{1+3u^2}du
\end{align*}
Both integrals are standard:
$$=-x+\frac{10}{\sqrt{3}}\arctan\left(\sqrt{3}u\right)+C$$
Undoing the substitution(s):
$$=-x+\frac{10}{\sqrt{3}}\arctan\left(\sqrt{3}\tan\left(\frac{x}{2}\right)\right)+C$$
\subsection*{Problem 73}
$$\int x^5\sqrt{x^3-1}dx$$
Rewriting the integral:
$$=\int x^2\cdot x^3\sqrt{x^3-1}dx$$
Substituting
\begin{align*}
u&=x^3-1 \\
u+1&=x^3 \\
du&=3x^2dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{3}\int(u+1)\sqrt{u}du \\
&=\frac{1}{3}\int\left(u^{3/2}+\sqrt{u}\right)du \\
&=\frac{1}{3}\int u^{3/2}du+\frac{1}{3}\int\sqrt{u}du
\end{align*}
Both integrals are standard:
$$=\frac{2}{15}u^{5/2}+\frac{2}{9}u^{3/2}+C$$
Undoing the substitution(s):
$$=\frac{2}{15}(x^3-1)^{5/2}+\frac{2}{9}(x^3-1)^{3/2}+C$$
\subsection*{Problem 74}
$$\int\frac{1}{2+2\cos\left(x\right)+\sin\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1}{2+2\left(\frac{1-u^2}{1+u^2}\right)+\frac{2u}{1+u^2}}\cdot\frac{2}{1+u^2}du \\
&=2\int\frac{1}{2(1+u^2)+2(1-u^2)+2u}du \\
&=2\int\frac{1}{2+2u^2+2-2u^2+2u}du \\
&=2\int\frac{1}{4+2u}du \\
&=\int\frac{1}{2+u}du
\end{align*}
Substituting
\begin{align*}
v&=2+u \\
dv&=du
\end{align*}
yields:
$$=\int\frac{1}{v}dv$$
This integral is standard:
$$=\ln|v|+C$$
Undoing the substitution(s):
$$=\ln\left|2+\tan\left(\frac{x}{2}\right)\right|+C$$
\subsection*{Problem 75}
$$\int\frac{\sqrt{1+\sin\left(x\right)}}{\sec\left(x\right)}dx$$
Rewriting the integral:
$$=\int\cos(x)\sqrt{1+\sin(x)}dx$$
Substituting
\begin{align*}
u&=1+\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\sqrt{u}du$$
This integral is standard:
$$=\frac{2}{3}u^{3/2}+C$$
Undoing the substitution(s):
$$=\frac{2}{3}(1+\sin(x))^{3/2}+C$$
\subsection*{Problem 76}
$$\int\frac{1}{x^{2/3}\left(1+x^{2/3}\right)}dx$$
Substituting
\begin{align*}
u&=\sqrt[3]{x} \\
du&=\frac{1}{3x^{2/3}}dx
\end{align*}
yields:
$$=3\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=3\arctan(u)+C$$
Undoing the substitution(s):
$$=3\arctan\left(\sqrt[3]{x}\right)+C$$
\subsection*{Problem 77}
$$\int\frac{\sin\left(x\right)}{\sin\left(2x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\sin(x)}{2\sin(x)\cos(x)}dx \\
&=\frac{1}{2}\int\frac{1}{\cos(x)}dx \\
&=\frac{1}{2}\int\sec(x)dx
\end{align*}
This integral is standard:
$$=\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C$$
\subsection*{Problem 78}
$$\int\sqrt{1+\cos\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{1+\cos(x)}\cdot\frac{\sqrt{1-\cos(x)}}{\sqrt{1-\cos(x)}}dx \\
&=\int\frac{\sqrt{1-\cos^2(x)}}{\sqrt{1-\cos(x)}}dx \\
&=\int\frac{\sqrt{\sin^2(x)}}{\sqrt{1-\cos(x)}}dx \\
&=\int\frac{\sin(x)}{\sqrt{1-\cos(x)}}dx
\end{align*}
Substituting
\begin{align*}
u&=1-\cos(x) \\
du&=\sin(x)dx
\end{align*}
yields:
$$=\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=2\sqrt{u}+C$$
Undoing the substitution(s):
$$=2\sqrt{1-\cos(x)}+C$$
\subsection*{Problem 79}
$$\int\sqrt{1+\sin\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{1+\sin(x)}\cdot\frac{\sqrt{1-\sin(x)}}{\sqrt{1-\sin(x)}}dx \\
&=\int\frac{1-\sin^2(x)}{\sqrt{1-\sin(x)}}dx \\
&=\int\frac{\sqrt{\cos^2(x)}}{\sqrt{1-\sin(x)}}dx \\
&=\int\frac{\cos(x)}{\sqrt{1-\sin(x)}}dx
\end{align*}
Substituting
\begin{align*}
u&=1-\sin(x) \\
du&=-\cos(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=-2\sqrt{u}+C$$
\subsection*{Problem 80}
$$\int\frac{\sec^2\left(x\right)}{1-\tan^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int\frac{1}{1-u^2}du$$
This integral is standard:
$$=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\ln\left|\frac{1+\tan(x)}{1-\tan(x)}\right|+C$$
\subsection*{Problem 81}
$$\int\ln\left(x^2+x+1\right)dx$$
Using integration by parts where
\begin{align*}
u&=\ln(x^2+x+1) &v&=x \\
du&=\frac{2x+1}{x^2+x+1}dx &dv&=dx
\end{align*}
yields:
$$=x\ln(x^2+x+1)-\int\frac{2x^2+x}{x^2+x+1}dx$$
Performing long division on the integral yields:
\begin{align*}
&=-\int\left(2+\frac{-x-2}{x^2+x+1}\right)dx \\
&=-2\int dx+\int\frac{x+2}{x^2+x+1}dx \\
&=-2\int dx+\int\frac{x+\frac{1}{2}+\frac{3}{2}}{x^2+x+1}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int\frac{1}{x^2+x+1}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int\frac{1}{x^2+x+\frac{1}{4}+\frac{3}{4}}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+6\int\frac{1}{4x^2+4x+1+3}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+6\int\frac{1}{(2x+1)^2+3}dx
\end{align*}
Substituting
\begin{align*}
w&=x^2+x+1 &z&=2x+1 \\
dw&=(2x+1)dx &dz&=2dx
\end{align*}
yields:
$$=-2\int dx+\frac{1}{2}\int\frac{1}{w}dw+3\int\frac{1}{z^2+3}$$
All three integrals are standard:
$$=-2x+\frac{1}{2}\ln|w|+\frac{3}{\sqrt{3}}\arctan\left(\frac{z}{\sqrt{3}}\right)+C$$
Undoing the substitution(s):
$$=x\ln(x^2+x+1)-2x+\frac{1}{2}\ln\left|x^2+x+1\right|+\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+C$$
\subsection*{Problem 82}
$$\int e^x\arcsin\left(e^x\right)dx$$
Substituting
\begin{align*}
u&=e^x \\
du&=e^xdx
\end{align*}
yields:
$$=\int\arcsin(u)du$$
This integral is standard:
$$=u\arcsin(u)+\sqrt{1-u^2}+C$$
Undoing the substitution(s):
$$=e^x\arcsin\left(e^x\right)+\sqrt{1-e^{2x}}+C$$
\subsection*{Problem 83}
$$\int\frac{\arctan\left(x\right)}{x^2}dx$$
Using integration by parts where:
\begin{align*}
u&=\arctan(x) &v&=-\frac{1}{x} \\
du&=\frac{1}{1+x^2}dx &dv&=\frac{1}{x^2}dx
\end{align*}
yields:
$$=-\frac{\arctan(x)}{x}+\int\frac{1}{x(1+x^2)}dx$$
Using partial fraction decomposition
$$=\int\left(\frac{A}{x}+\frac{Bx+C}{1+x^2}\right)dx$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx \\
&=\int\frac{1}{x}dx-\int\frac{x}{x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=x^2+1 \\
du&=2xdx
\end{align*}
yields:
$$=-\frac{1}{2}\int\frac{1}{u}du$$
This integral is standard:
$$=-\frac{1}{2}\ln|u|+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{\arctan(x)}{x}+\ln|x|-\frac{1}{2}\ln\left|x^2+1\right|+C \\
&=-\frac{\arctan(x)}{x}+\ln\left|\frac{x}{\sqrt{x^2+1}}\right|+C
\end{align*}
\subsection*{Problem 84}
$$\int\frac{x^2}{\sqrt{x^2-25}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^2}{\sqrt{25\left(\frac{x^2}{25}-1\right)}}dx \\
&=\frac{1}{5}\int\frac{x^2}{\sqrt{\left(\frac{x}{5}\right)^2-1}}dx
\end{align*}
Substituting
\begin{align*}
\sec\theta&=\frac{x}{5} \\
5\sec\theta&=x \\
5\sec\theta\tan\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{(5\sec\theta)^2}{\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta d\theta \\
&=25\int\frac{\sec^3\theta\tan\theta}{\sqrt{\tan^2\theta}}d\theta \\
&=25\int\sec^3\theta d\theta
\end{align*}
This integral is standard:
\begin{align*}
&=25\left(\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C\right) \\
&=\frac{25}{2}\sec\theta\tan\theta+\frac{25}{2}\ln\left|\sec\theta+\tan\theta\right|+C
\end{align*}
Undoing the substitution(s):
\begin{align*}
&=\frac{25}{2}\cdot\frac{x}{5}\cdot\frac{\sqrt{x^2-25}}{5}+\frac{25}{2}\ln\left|\frac{x}{5}+\frac{\sqrt{x^2-25}}{5}\right|+C \\
&=\frac{1}{2}x\sqrt{x^2-25}+\frac{25}{2}\ln\left|x+\sqrt{x^2-25}\right|+C
\end{align*}
\subsection*{Problem 85}
$$\int\frac{x^3}{\left(x^2+1\right)^2}dx$$
Rewriting the integral:
$$=\int\frac{x\cdot x^2}{(x^2+1)^2}dx$$
Substituting
\begin{align*}
u&=x^2+1 \\
u-1&=x^2 \\
du&=2xdx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{2}\int\frac{u-1}{u^2}du \\
&=\frac{1}{2}\int\left(\frac{1}{u}-\frac{1}{u^2}\right)du \\
&=\frac{1}{2}\int\frac{1}{u}du-\frac{1}{2}\int\frac{1}{u^2}du
\end{align*}
All three integrals are standard:
$$=\frac{1}{2}\ln|u|-\frac{1}{2u}+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\ln\left|x^2+1\right|+\frac{1}{2x^2+2}+C$$
\subsection*{Problem 86}
$$\int\frac{1}{x\sqrt{6x-x^2}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{1}{x\sqrt{9-9+6x-x^2}}dx \\
&=\int\frac{1}{x\sqrt{9-(9-6x+x^2)}}dx \\
&=\int\frac{1}{x\sqrt{9-(3-x)^2}}dx \\
&=\int\frac{1}{x\sqrt{9\left(1-\frac{(3-x)^2}{9}\right)}}dx \\
&=\frac{1}{3}\int\frac{1}{x\sqrt{1-\left(\frac{3-x}{3}\right)^2}}dx
\end{align*}
Substituting
\begin{align*}
\sin\theta&=\frac{3-x}{3} \\
3-3\sin\theta&=x \\
-3\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=-\int\frac{\cos\theta}{(3-3\sin\theta)\sqrt{1-\sin^2\theta}}d\theta \\
&=-\int\frac{\cos\theta}{(3-3\sin\theta)\sqrt{\cos^2\theta}}d\theta \\
&=-\int\frac{1}{3-3\sin\theta}d\theta \\
&=-\frac{1}{3}\int\frac{1}{1-\sin\theta}\cdot\frac{1+\sin\theta}{1+\sin\theta}d\theta \\
&=-\frac{1}{3}\int\frac{1+\sin\theta}{1-\sin^2\theta}d\theta \\
&=-\frac{1}{3}\int\frac{1+\sin\theta}{\cos^2\theta}d\theta \\
&=-\frac{1}{3}\int\left(\sec^2\theta+\sec\theta\tan\theta\right)d\theta \\
&=-\frac{1}{3}\int\sec^2\theta d\theta-\frac{1}{3}\int\sec\theta\tan\theta d\theta
\end{align*}
Both integrals are standard:
$$=-\frac{1}{3}\tan\theta-\frac{1}{3}\sec\theta+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{1}{3}\cdot\frac{3-x}{\sqrt{6x-x^2}}-\frac{1}{3}\cdot\frac{3}{\sqrt{6x-x^2}}+C \\
&=\frac{x-3}{3\sqrt{6x-x^2}}-\frac{3}{3\sqrt{6x-x^2}}+C \\
&=\frac{x-6}{3\sqrt{6x-x^2}}+C
\end{align*}
\subsection*{Problem 87}
$$\int\frac{3x+2}{\left(x^2+4\right)^{3/2}}dx$$
Rewriting the integral:
$$\frac{3}{2}\int\frac{2x}{(x^2+4)^{3/2}}dx+\frac{1}{4}\int\frac{1}{\left(\left(\frac{x}{2}\right)^2+1\right)^{3/2}}dx$$
Substituting
\begin{align*}
u&=x^2+4 &\tan\theta&=\frac{x}{2} \\
du&=2xdx &\sec^2\theta d\theta&=\frac{1}{2}dx
\end{align*}
yields:
\begin{align*}
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\frac{\sec^2\theta}{(\tan^2\theta+1)^{3/2}}d\theta \\
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\frac{\sec^2\theta}{(\sec^2\theta)^{3/2}}d\theta \\
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\frac{1}{\sec\theta}d\theta \\
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\cos\theta d\theta
\end{align*}
Both integrals are standard:
$$=-\frac{3}{\sqrt{u}}+\frac{1}{2}\sin\theta+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{3}{\sqrt{x^2+4}}+\frac{1}{2}\cdot\frac{x}{\sqrt{x^2+4}}+C \\
&=\frac{x-6}{2\sqrt{x^2+4}}+C
\end{align*}
\subsection*{Problem 88}
$$\int x^{3/2}\ln\left(x\right)dx$$
Using integration by parts where
\begin{align*}
u&=\ln(x) &v&=\frac{2}{5}x^{5/2} \\
du&=\frac{1}{x}dx &dv&=x^{3/2}dx
\end{align*}
yields:
$$=\frac{2}{5}x^{5/2}\ln(x)-\frac{2}{5}\int x^{3/2}dx$$
This integral is standard:
$$=\frac{2}{5}x^{5/2}\ln(x)-\frac{4}{25}x^{5/2}+C$$
\subsection*{Problem 89}
$$\int\frac{\sqrt{1+\sin^2\left(x\right)}}{\sec\left(x\right)\csc\left(x\right)}dx$$
Rewriting the integral;
$$=\int\sin(x)\cos(x)\sqrt{1+\sin^2(x)}dx$$
Substituting
\begin{align*}
u&=1+\sin^2(x) \\
du&=2\sin(x)\cos(x)dx
\end{align*}
yields:
$$=\frac{1}{2}\int\sqrt{u}du$$
This integral is standard:
$$=\frac{1}{3}u^{3/2}+C$$
Undoing the substitution(s):
$$=\frac{1}{3}\left(1+\sin^2(x)\right)^{3/2}+C$$
\subsection*{Problem 90}
$$\int\frac{e^{\sqrt{\sin\left(x\right)}}}{\sec\left(x\right)\sqrt{\sin\left(x\right)}}dx$$
Rewriting the integral:
$$=\int\frac{\cos(x)e^{\sqrt{\sin(x)}}}{\sqrt{\sin(x)}}dx$$
Substituting
\begin{align*}
u&=\sqrt{\sin(x)} \\
du&=\frac{\cos(x)}{2\sqrt{\sin(x)}}dx
\end{align*}
yields
$$=2\int e^udu$$
This integral is standard:
$$=2e^u+C$$
Undoing the substitution(s):
$$=2e^{\sqrt{\sin(x)}}+C$$
\subsection*{Problem 91}
$$\int xe^x\sin(x)dx$$
Recall that
$$\int e^x\sin(x)dx=\frac{1}{2}e^x\sin(x)-\frac{1}{2}e^x\cos(x)$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| } 
 \hline
 Sign & D & I \\
 \hline
 $+$ & $xe^x$ & $\sin(x)$ \\ 
 $-$ & $e^x+xe^x$ & $-\cos(x)$ \\
 $+$ & $2e^x+xe^x$ & $-\sin(x)$ \\
 \hline
\end{tabular}
\end{center}
yields:
\begin{align*}
&=-xe^x\cos(x)+e^x\sin(x)(1+x)-\int e^x\sin(x)(2+x)dx \\
2\int xe^x\sin(x)dx&=-xe^x\cos(x)+e^x\sin(x)(1+x)-2\int e^x\sin(x)dx \\
&=-xe^x\cos(x)+xe^x\sin(x)+e^x\sin(x)-e^x\sin(x)+e^x\cos(x) \\
\int xe^x\sin(x)dx&=\frac{1}{2}xe^x(\sin(x)-\cos(x))+\frac{1}{2}e^x\cos(x)+C
\end{align*}
\subsection*{Problem 92}
$$\int x^2e^{x^{3/2}}dx$$
Rewriting the integral:
$$=\int x^{3/2}\cdot x^{1/2}e^{x^{3/2}}dx$$
Substituting
\begin{align*}
u&=x^{3/2} \\
du&=\frac{3}{2}x^{1/2}dx
\end{align*}
yields:
$$=\frac{2}{3}\int ue^udu$$
Using integration by parts where
\begin{align*}
w&=u &v&=e^u \\
dw&=du &dv&=e^udu
\end{align*}
yields:
$$=\frac{2}{3}ue^u-\frac{2}{3}\int e^udu$$
This integral is standard:
$$=\frac{2}{3}ue^u-\frac{2}{3}e^u+C$$
Undoing the substitution(s):
$$=\frac{2}{3}x^{3/2}e^{x^{3/2}}-\frac{2}{3}e^{x^{3/2}}+C$$
\subsection*{Problem 93}
$$\int\frac{\arctan\left(x\right)}{\left(x-1\right)^3}dx$$
Using integration by parts where
\begin{align*}
u&=\arctan(x) &v&=-\frac{1}{2(x-1)^2}dx \\
du&=\frac{1}{1+x^2}dx &dv&=\frac{1}{(x-1)^3}dx
\end{align*}
yields:
$$=-\frac{\arctan(x)}{2(x-1)^2}+\frac{1}{2}\int\frac{1}{(1+x^2)(x-1)^2}dx$$
Using partial fraction decomposition
$$=\frac{1}{2}\int\left(\frac{Ax+B}{1+x^2}+\frac{C}{(x-1)^2}+\frac{D}{x-1}\right)dx$$
where $A,B,C,D\in\mathbb{R}$ yields:
\begin{align*}
&=\frac{1}{2}\bigintss\left(\frac{\frac{1}{2}x+0}{x^2+1}+\frac{-\frac{1}{2}}{x-1}+\frac{\frac{1}{2}}{\left(x-1\right)^2}\right)dx \\
&=\frac{1}{4}\int\left(\frac{x}{x^2+1}-\frac{1}{x-1}+\frac{1}{(x-1)^2}\right)dx \\
&=\frac{1}{4}\int\frac{x}{x^2+1}dx-\frac{1}{4}\int\frac{1}{x-1}dx+\frac{1}{4}\int\frac{1}{(x-1)^2}dx
\end{align*}
Substituting
\begin{align*}
w_1&=x^2+1 &w_2&=x-1 \\
dw_1&=2xdx &dw_2&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{8}\int\frac{1}{w_1}dw_1-\frac{1}{4}\int\frac{1}{w_2}dw_2+\frac{1}{4}\int\frac{1}{w_2^2}dw_2
\end{align*}
All three integrals are standard:
\begin{align*}
&=\frac{1}{8}\ln\left|w_1\right|-\frac{1}{4}\ln\left|w_2\right|-\frac{1}{4w_2} \\
&=\frac{1}{8}\ln\left|\frac{w_1}{w_2^2}\right|-\frac{1}{4w_2}
\end{align*}
Undoing the substitution(s):
$$=-\frac{\arctan(x)}{2(x-1)^2}+\frac{1}{8}\ln\left|\frac{x^2+1}{(x-1)^2}\right|-\frac{1}{4(x-1)}+C$$
\subsection*{Problem 94}
$$\int\ln\left(1+\sqrt{x}\right)dx$$
Substituting
\begin{align*}
u&=1+\sqrt{x} \\
u-1&=\sqrt{x} \\
u^2-2u+1&=x \\
(2u-2)du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\ln(u)(2u-2)du \\
&=2\int u\ln(u)du-2\int\ln(u)du
\end{align*}
Using integration by parts where
\begin{align*}
w&=\ln(u) &v&=\frac{1}{2}u^2 \\
dw&=\frac{1}{u}du &dv&=udu
\end{align*}
yields:
\begin{align*}
&=2\left(\frac{1}{2}u^2\ln(u)-\frac{1}{2}\int udu\right) \\
&=u^2\ln(u)-\int udu
\end{align*}
The remaining two integrals are standard:
\begin{align*}
&=u^2\ln(u)-\frac{1}{2}u^2-2u\ln(u)+2u+C \\
&=u\ln(u)\left(u-2\right)-\frac{1}{2}u^2+2u+C
\end{align*}
Undoing the substitution(s):
$$=\left(1+\sqrt{x}\right)\ln\left(1+\sqrt{x}\right)\left(\sqrt{x}-1\right)-\frac{1}{2}\left(1+\sqrt{x}\right)^2+2\sqrt{x}+C$$
\subsection*{Problem 95}
$$\int\frac{2x+3}{\sqrt{3+6x-9x^2}}dx$$
Rewriting the integral:
\begin{align*}
&=\bigintss\frac{2x-\frac{2}{3}+\frac{11}{3}}{\sqrt{3+6x-9x^2}}dx \\
&=-\frac{1}{9}\int\frac{6-18x}{\sqrt{3+6x-9x^2}}+\frac{11}{3}\int\frac{1}{\sqrt{4-1+6x-9x^2}}dx \\
&=-\frac{1}{9}\int\frac{6-18x}{\sqrt{3+6x-9x^2}}+\frac{11}{3}\int\frac{1}{\sqrt{4-(1-6x+9x^2)}}dx \\
&=-\frac{1}{9}\int\frac{6-18x}{\sqrt{3+6x-9x^2}}+\frac{11}{3}\int\frac{1}{\sqrt{4-(1-3x)^2}}dx
\end{align*}
Substituting
\begin{align*}
u&=3+6x-9x^2 &v&=1-3x \\
du&=(6-18x)dx &dv&=-3dx
\end{align*}
yields:
$$=-\frac{1}{9}\int\frac{1}{\sqrt{u}}du-\frac{11}{9}\int\frac{1}{\sqrt{4-v^2}}dv$$
Both integrals are standard:
$$=-\frac{2}{9}\sqrt{u}-\frac{11}{9}\arcsin\left(\frac{v}{2}\right)+C$$
Undoing the substitution(s):
$$=-\frac{2}{9}\sqrt{3+6x-9x^2}-\frac{11}{9}\arcsin\left(\frac{1-3x}{2}\right)+C$$
\subsection*{Problem 96}
$$\int\frac{1}{2+2\sin\left(x\right)+\cos\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\bigintss\frac{1}{2+2\left(\frac{2u}{1+u^2}\right)+\left(\frac{1-u^2}{1+u^2}\right)}\cdot\frac{2}{1+u^2}du \\
&=2\int\frac{1}{2+2u^2+4u+1-u^2}du \\
&=2\int\frac{1}{u^2+4u+3}du \\
&=2\int\frac{1}{(u+3)(u+1)}du \\
&=2\bigintss\left(\frac{-\frac{1}{2}}{u+3}+\frac{\frac{1}{2}}{u+1}\right)du \\
&=-\int\frac{1}{u+3}du+\int\frac{1}{u+1}du
\end{align*}
Substituting
\begin{align*}
v_1&=u+3 &v_2&=u+1 \\
dv_1&=du &dv_2&=du
\end{align*}
yields:
$$=-\int\frac{1}{v_1}dv_1+\int\frac{1}{v_2}dv_2$$
Both integrals are standard:
\begin{align*}
&=-\ln|v_1|+\ln|v_2|+C \\
&=\ln\left|\frac{v_2}{v_1}\right|+C
\end{align*}
Undoing the substitution(s):
$$\ln\left|\frac{\tan\left(\frac{x}{2}\right)+1}{\tan\left(\frac{x}{2}\right)+3}\right|+C$$
\subsection*{Problem 97}
$$\int\frac{\sin^3\left(x\right)}{\cos\left(x\right)-1}dx$$
Rewriting the integral:
\begin{align*}
&=-\int\frac{\sin(x)\cdot\sin^2(x)}{1-\cos(x)}dx \\
&=-\int\frac{\sin(x)(1-\cos^2(x))}{1-\cos(x)}dx \\
&=-\int\frac{\sin(x)(1-\cos(x))(1+\cos(x))}{1-\cos(x)}dx \\
&=-\int\sin(x)(1+\cos(x))dx
\end{align*}
Substituting
\begin{align*}
u&=1+\cos(x)\\
du&=-\sin(x)dx
\end{align*}
yields:
$$=\int udu$$
This integral is standard:
$$=\frac{1}{2}u^2+C$$
Undoing the substitution(s):
$$=\frac{1}{2}(1+\cos(x))^2+C$$
\subsection*{Problem 98}
$$\int x^{3/2}\arctan\left(\sqrt{x}\right)dx$$
Using integration by parts where:
\begin{align*}
u&=\arctan\left(\sqrt{x}\right) &v&=\frac{2}{5}x^{5/2} \\
dw&=\frac{1}{2\sqrt{x}(1+x)}dx &dv&=x^{3/2}dx
\end{align*}
yields:
\begin{align*}
&=\frac{2}{5}x^{5/2}\arctan\left(\sqrt{x}\right)-\frac{1}{5}\int\frac{x^2}{1+x}dx
\end{align*}
Substituting
\begin{align*}
w&=1+x \\
w-1&=x \\
dw&=dx
\end{align*}
yields:
\begin{align*}
&=-\frac{1}{5}\int\frac{(w-1)^2}{w}dw \\
&=-\frac{1}{5}\int\frac{w^2-2w+1}{w}dw \\
&=-\frac{1}{5}\int\left(w-2+\frac{1}{w}\right)dw \\
&=-\frac{1}{5}\int wdw+\frac{2}{5}\int dw-\frac{1}{5}\int\frac{1}{w}dw
\end{align*}
All three integrals are standard:
$$-\frac{1}{10}w^2+\frac{2}{5}w-\frac{1}{5}\ln|w|+C$$
Undoing the substitution(s):
$$=\frac{2}{5}x^{5/2}\arctan\left(\sqrt{x}\right)-\frac{1}{10}(1+x)^2+\frac{2}{5}(1+x)-\frac{1}{5}\ln|1+x|+C$$
\subsection*{Problem 99}
$$\int\arcsec\left(\sqrt{x}\right)dx$$
Substituting
\begin{align*}
u&=\sqrt{x} \\
u^2&=x \\
2udu&=dx
\end{align*}
yields:
$$=2\int u\arcsec(u)du$$
Using integration by parts where:
\begin{align*}
w&=\arcsec(u) &v&=\frac{1}{2}u^2 \\
dw&=\frac{1}{u\sqrt{u^2-1}}du &dv&=udu
\end{align*}
yields:
\begin{align*}
&=2\left(\frac{1}{2}u^2\arcsec(u)-\frac{1}{2}\int\frac{u^2}{u\sqrt{u^2-1}}du\right) \\
&=u^2\arcsec(u)-\int\frac{u}{\sqrt{u^2-1}}du
\end{align*}
Substituting
\begin{align*}
z&=u^2-1 \\
dz&=2udu
\end{align*}
yields:
$$=u^2\arcsec(u)-\frac{1}{2}\int\frac{1}{\sqrt{z}}dz$$
This integral is standard:
$$=u^2\arcsec(u)-\sqrt{z}+C$$
Undoing the substitution(s):
\begin{align*}
&=x\arcsec\left(\sqrt{x}\right)-\sqrt{x-1}+C
\end{align*}
\subsection*{Problem 100}
$$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$$
Substituting
\begin{align*}
\sin\theta&=x^2 \\
\cos\theta d\theta&=2xdx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{1-\sin\theta}{1+\sin\theta}\cdot\frac{1-\sin\theta}{1-\sin\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\cdot\frac{1-\sin\theta}{\cos\theta}d\theta \\
&=\frac{1}{2}\int(1-\sin\theta)d\theta \\
&=\frac{1}{2}\int d\theta-\frac{1}{2}\int\sin\theta d\theta
\end{align*}
Both integrals are standard:
$$=\frac{1}{2}\theta+\frac{1}{2}\cos\theta+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\arcsin\left(x^2\right)+\frac{1}{2}\sqrt{1-x^4}+C$$
\newpage
\section{Advanced Integrals - Proofs}
\subsection*{Advanced Integral 1}
$$\int\sec^3(x)dx$$
Rewriting the integral:
$$=\int\sec(x)\sec^2(x)dx$$
Using integration by parts where
\begin{align*}
u&=\sec(x) &v&=\tan(x) \\
du&=\sec(x)\tan(x)dx &dv&=\sec^2(x)dx
\end{align*}
yields:
\begin{align*}
&=\sec(x)\tan(x)-\int\sec(x)\tan^2(x)dx \\
&=\sec(x)\tan(x)-\int\sec(x)(\sec^2(x)-1)dx \\
&=\sec(x)\tan(x)-\int(\sec^3(x)-\sec(x))dx \\
&=\sec(x)\tan(x)-\int\sec^3(x)dx+\int\sec(x)dx \\
2\int\sec^3(x)dx&=\sec(x)\tan(x)+\ln\left|\sec(x)+\tan(x)\right|+C \\
&=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C
\end{align*}
\subsection*{Advanced Integral 2}
$$\int\arcsin(x)dx$$
Using integration by parts where
\begin{align*}
u&=\arcsin(x) &v&=x \\
du&=\frac{1}{\sqrt{1-x^2}}dx &dv&=dx
\end{align*}
yields:
$$x\arcsin(x)-\int\frac{x}{\sqrt{1-x^2}}dx$$
Substituting
\begin{align*}
u&=1-x^2 \\
du&=-2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=\sqrt{u}+C$$
Undoing the substitution(s):
$$=x\arcsin(x)+\sqrt{1-x^2}+C$$
\subsection*{Advanced Integral 3}
$$\int\arccos(x)dx$$
Using integration by parts where
\begin{align*}
u&=\arccos(x) &v&=x \\
du&=-\frac{1}{\sqrt{1-x^2}}dx &dv&=dx
\end{align*}
yields:
$$x\arccos(x)+\int\frac{x}{\sqrt{1-x^2}}dx$$
Substituting
\begin{align*}
u&=1-x^2 \\
du&=-2xdx
\end{align*}
yields:
$$=-\frac{1}{2}\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=-\sqrt{u}+C$$
Undoing the substitution(s):
$$=x\arccos(x)-\sqrt{1-x^2}+C$$
\subsection*{Advanced Integral 4}
$$\int\arctan(x)dx$$
Using integration by parts where
\begin{align*}
u&=\arctan(x) &v&=x \\
du&=\frac{1}{1+x^2}dx &dv&=dx
\end{align*}
yields:
$$=x\arctan(x)-\int\frac{x}{1+x^2}dx$$
Substituting
\begin{align*}
u&=1+x^2 \\
du&=2xdx
\end{align*}
yields:
$$=-\frac{1}{2}\int\frac{1}{u}du$$
This integral is standard:
$$=x\arctan(x)-\frac{1}{2}\ln\left|1+x^2\right|+C$$
\subsection*{Advanced Integral 5}
$$\int\sqrt{x^2+a^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{a^2\left(\frac{x^2}{a^2}+1\right)}dx \\
&=a\int\sqrt{\left(\frac{x}{a}\right)^2+1}dx
\end{align*}
Substituting
$$$$
\subsection*{Advanced Integral 6}
$$\int\sqrt{x^2-a^2}dx$$
\subsection*{Advanced Integral 7}
$$\int\sqrt{a^2-x^2}dx$$
\subsection*{Advanced Integral 8}
$$\int\frac{1}{x^2+a^2}dx$$
\subsection*{Advanced Integral 9}
$$\int\frac{1}{x^2-a^2}dx$$
\subsection*{Advanced Integral 10}
$$\int\frac{1}{a^2-x^2}dx$$
\subsection*{Advanced Integral 11}
$$\int\frac{1}{\sqrt{x^2+a^2}}dx$$
\subsection*{Advanced Integral 12}
$$\int\frac{1}{\sqrt{x^2-a^2}}dx$$
\subsection*{Advanced Integral 13}
$$\int\frac{1}{\sqrt{a^2-x^2}}dx$$
\subsection*{Advanced Integral 14}
$$\int\sin^n(x)dx$$
\subsection*{Advanced Integral 15}
$$\int\cos^n(x)dx$$
\end{document}