# Integration Tables

$\DeclareMathOperator{\arccot}{arccot}$

$\int {u\mbox{ }dv} =uv-\int {v\mbox{ }du}$

$\int {u^{n}\mbox{ }d} u=\frac{1}{n+1}u^{n+1}+C,\mbox{n}\ne \mbox{-1}$

$\int {\frac{du}{u}} =\ln \left| u \right|+C$

$\int {e^{u}du=e^{u}+C}$

$\int {e^{u}du=e^{u}+C}$

$\int {\sin \left( u \right)du=-\cos \left( u \right)+C}$

$\int {\cos \left( u \right)du=\sin \left( u \right)+C}$

$\int {\sec^{2}\left( u \right)du=\tan \left( u \right)+C}$

$\int {\csc \left( u \right)\cot \left( u \right)du=-\csc \left( u \right)+C}$

$\int {\tan \left( u \right)du=\ln \left| {\sec \left( u \right)} \right|} +C$

$\int {\cot \left( u \right)du=\ln \left| {\sin \left( u \right)} \right|+C}$

$\int {\sec \left( u \right)du=\ln \left| {\sec \left( u \right)+\tan \left( u \right)} \right|} +C$

$\int {\csc \left( u \right)du=\ln \left| {\csc \left( u \right)-\cot \left( u \right)} \right|+C}$

$\int {\frac{du}{\sqrt {a^{2}-u^{2}} }=\sin^{-1}\left( {\frac{u}{a}} \right)+C}$

$\int {\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\left( {\frac{u}{a}} \right)+C}$

$\int {\frac{du}{u\sqrt {a^{2}-a^{2}} }=\frac{1}{a}\sec^{-1}\left( {\frac{u}{a}} \right)} +C$

$\int {\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\ln \left| {\frac{u+a}{u-a}} \right|} +C$

$\int {\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\ln \left| {\frac{u+a}{u-a}} \right|} +C$

$\int {\frac{du}{u^{2}-a^{2}}=\frac{1}{2a}\ln \left| {\frac{u-a}{u+a}} \right|} +C$

$\int {\frac{u}{a+bu}du=\frac{1}{b^{2}}\left( {a+bu-a\ln \left| {a+bu} \right|} \right)+C}$

$\int {\frac{u^{2}}{a+bu}du=\frac{1}{2b^{3}}\left[ {\left( {a+bu} \right)^{2}-4a\left( {a+bu} \right)+2a^{2}\ln \left| {a+bu} \right|} \right]+C}$

$\int {\frac{du}{u\left( {a+bu} \right)}=\frac{1}{a}\ln \left| {\frac{u}{a+bu}} \right|+C}$

$\int {\frac{du}{u^{2}\left( {a+bu} \right)}=-\frac{1}{au}+\frac{b}{a^{2}}\ln \left| {\frac{a+bu}{u}} \right|+C}$

$\int {\frac{du}{\left( {a+bu} \right)^{2}}=\frac{a}{b^{2}\left( {a+bu} \right)}+\frac{1}{b^{2}}\ln \left| {a+bu} \right|+C}$

$\int {\frac{du}{u\left( {a+bu} \right)^{2}}=\frac{1}{a\left( {a+bu} \right)}+\frac{1}{a^{2}}\ln \left| {\frac{a+bu}{u}} \right|+C}$

$\int {\frac{u^{2}}{\left( {a+bu} \right)^{2}}du=\frac{1}{b^{3}}\left( {a+bu-\frac{a^{2}}{a+bu}-2a\ln \left| {a+bu} \right|} \right)+C}$

$\int {u\sqrt {a+bu} du=\frac{2}{15b^{2}}\left( {3bu-2a} \right)\left( {a+bu} \right)^{\frac{3}{2}}+C}$

$\int {\frac{u}{\sqrt {a+bu} } du=\frac{2}{3b^{2}}\left( {bu-2a} \right)\sqrt {a+bu} +C}$

$\int {\frac{u^{2}}{\sqrt {a+bu} } du=\frac{2}{15b^{3}}\left( {8a^{2}+3b^{2}u^{2}-4abu} \right)\sqrt {a+bu} +C}$

\begin{aligned} \int {\frac{du}{u\sqrt {a+bu} }} &=\frac{1}{\sqrt a }\ln \left| {\frac{\sqrt {a+bu} -\sqrt a }{\sqrt {a+bu} +\sqrt a }} \right|+C\mbox{, if a}>\mbox{0} \\ &=\frac{2}{\sqrt {-a} }\tan^{-1}\sqrt {\frac{a+bu}{-a}} +C\mbox{, if a}<\mbox{0} \\ \end{aligned}

$\int {\frac{\sqrt {a+bu} }{\mbox{u}} du=2\sqrt {a+bu} +a\int {\frac{du}{u\sqrt {a+bu} }} }$

$\int {\frac{\sqrt {a+bu} }{\mbox{u}^{\mbox{2}}} du=-\frac{\sqrt {a+bu} }{u}+\frac{b}{2}\int {\frac{du}{u\sqrt {a+bu} }} }$

$\int {u^{n}\sqrt {a+bu} du=\frac{2}{b\left( {2n+3} \right)}\left[ {u^{n}\left( {a+bu} \right)^{\frac{3}{2}}-na\int {u^{n-1}\sqrt {a+bu} du} } \right]}$

$\int {\frac{u^{n}}{\sqrt {a+bu} } du=\frac{2u^{n}\sqrt {a+bu} }{b\left( {2n+1} \right)}-\frac{2na}{b\left( {2n+a} \right)}} \int {\frac{u^{n-1}}{\sqrt {a+bu} } du}$

$\int {\frac{du}{u^{n}\sqrt {a+bu} } =-\frac{\sqrt {a+bu} }{a\left( {n-1} \right)u^{n-1}}-\frac{b\left( {2n-3} \right)}{2a\left( {n-1} \right)}} \int {\frac{du}{u^{n-1}\sqrt {a+bu} } }$

$\int {\sin^{2}\left( u \right)} du=\frac{1}{2}u-\frac{1}{4}\sin \left( {2u} \right)+C$

$\int {\cos^{2}\left( u \right)} du=\frac{1}{2}u+\frac{1}{4}\sin \left( {2u} \right)+C$

$\int {\tan^{2}\left( u \right) du=\tan \left( u \right)-u+C}$

$\int {\cot^{2}\left( u \right) du=-\cot \left( u \right)-u+C}$

$\int {\sin^{3}\left( u \right) du} =-\frac{1}{3}\left[ {2+\sin^{2}\left( u \right)} \right]\cos \left( u \right)+C$

$\int {\tan^{3}\left( u \right) du} =\frac{1}{2}\tan^{2}\left( u \right)+\ln \left| {\cos \left( u \right)} \right|+C$

$\int {\cot^{3}\left( u \right) du} =-\frac{1}{2}\cot^{2}\left( u \right)+\ln \left| {\sin \left( u \right)} \right|+C$

$\int {\sec^{3}\left( u \right) du} =\frac{1}{2}\sec \left( u \right)\tan \left( u \right)+\frac{1}{2}\ln \left| {\sec \left( u \right)+\tan \left( u \right)} \right|+C$

$\int {\csc^{3}\left( u \right) du} =\frac{1}{2}\csc \left( u \right)\cot \left( u \right)+\frac{1}{2}\ln \left| {\csc \left( u \right)+\cot \left( u \right)} \right|+C$

$\int {\sin^{n}\left( u \right) du=-\frac{1}{n}\sin^{n-1}\left( u \right)\cos \left( u \right)+} \frac{n-1}{n}\int {\sin^{n-2}\left( u \right) du} +C$

$\int {\cos^{n}\left( u \right) du=\frac{1}{n}\cos^{n-1}\left( u \right)\sin \left( u \right)+} \frac{n-1}{n}\int {\cos^{n-2}\left( u \right) du} +C$

$\int {\tan^{n}\left( u \right) du=\frac{1}{n-1}\tan^{n-1}\left( u \right)} -\int {\tan^{n-2}\left( u \right) du} +C$

$\int {\cot^{n}\left( u \right) du=\frac{-1}{n-1}\cot^{n-1}\left( u \right)} -\int {\cot^{n-2}\left( u \right) du} +C$

$\int {\sec^{n}\left( u \right) du=\frac{1}{n-1}\tan \left( u \right)\sec^{n-2}\left( u \right)} -\frac{n-2}{n-1}\int {\sec^{n-2}\left( u \right) du} +C$

$\int {\csc^{n}\left( u \right) du=\frac{-1}{n-1}\cot \left( u \right)\csc^{n-2}\left( u \right)} -\frac{n-2}{n-1}\int {\csc^{n-2}\left( u \right) du} +C$

$\int {\sin \left( {au} \right)\sin \left( {bu} \right) du=\frac{\sin \left( {a-b} \right)u}{2\left( {a-b} \right)}-\frac{\sin \left( {a+b} \right)u}{2\left( {a+b} \right)}+C}$

$\int {\cos \left( {au} \right)\cos \left( {bu} \right) du=\frac{\sin \left( {a-b} \right)u}{2\left( {a-b} \right)}+\frac{\sin \left( {a+b} \right)u}{2\left( {a+b} \right)}+C}$

$\int {\sin \left( {au} \right)\cos \left( {bu} \right) du=-\frac{\cos \left( {a-b} \right)u}{2\left( {a-b} \right)}-\frac{\sin \left( {a+b} \right)u}{2\left( {a+b} \right)}+C}$

$\int {u\sin \left( u \right) du=\sin \left( u \right)-u\cos \left( u \right)+C}$

$\int {u\cos \left( u \right) du=\cos \left( u \right)-u\sin \left( u \right)+C}$

$\int {u^{n}\sin \left( u \right) du=-u^{n}\cos \left( u \right)+n\int {u^{n-1}\cos \left( u \right) du} }$

$\int {u^{n}\cos \left( u \right) du=u^{n}\sin \left( u \right)-n\int {u^{n-1}\sin \left( u \right) du} }$

\begin{aligned} \int \sin^{n}\left( u \right)\cos^{m}\left( u \right) du &= -\frac{\sin^{n-1}\left( u \right)\cos^{m+1}\left( u \right)}{n+m}+\frac{n-1}{n+m}\int {\sin^{n-2}\left( u \right)\mbox{cos}^{\mbox{m}}\left( {\mbox{u}} \right) du} \\ &= \frac{\sin^{n+1}\left( u \right)\cos^{m-1}\left( u \right)}{n+m}+\frac{m-1}{n+m}\int {\sin^{n}\left( u \right)\mbox{cos}^{\mbox{m-2}}\left( {\mbox{u}} \right) du} \end{aligned}

$\int {\sin^{-1}\left( u \right) du} =u\sin^{-1}\left( u \right)+\sqrt {1-u^{2}} +C$

$\int {\cos^{-1}\left( u \right) du} =u\cos^{-1}\left( u \right)-\sqrt {1-u^{2}} +C$

$\int {\tan^{-1}\left( u \right) du} =u\tan^{-1}\left( u \right)-\frac{1}{2}\ln \left( {1+u^{2}} \right)+C$

$\int {u\sin^{-1}\left( u \right) du} =\frac{2u^{2}-1}{4}\sin^{-1}\left( u \right)+\frac{u\sqrt {1-u^{2}} }{4}+C$

$\int {u\cos^{-1}\left( u \right) du} =\frac{2u^{2}-1}{4}\cos^{-1}\left( u \right)-\frac{u\sqrt {1-u^{2}} }{4}+C$

$\int {u\tan^{-1}\left( u \right) du} =\frac{u^{2}+1}{2}\tan^{-1}\left( u \right)-\frac{u}{2}+C$

$\int {u^{n}\sin^{-1}\left( u \right) du} =\frac{1}{n+1}\left[ {u^{n+1}\sin^{-1}\left( u \right)-\int {\frac{u^{n+1}du}{\sqrt {1-u^{2}} }} } \right],n\ne -1$

$\int {u^{n}\cos^{-1}\left( u \right) du} =\frac{1}{n+1}\left[ {u^{n+1}\cos^{-1}\left( u \right)+\int {\frac{u^{n+1}du}{\sqrt {1-u^{2}} }} } \right],n\ne -1$

$\int {u^{n}\tan^{-1}\left( u \right) du} =\frac{1}{n+1}\left[ {u^{n+1}\tan^{-1}\left( u \right)-\int {\frac{u^{n+1}du}{\sqrt {1+u^{2}} }} } \right],n\ne -1$

$\int {ue^{au}du=\frac{1}{a^{2}}\left( {au-1} \right)e^{au}+C}$

$\int {u^{n}e^{au}du=\frac{1}{a}u^{n}e^{au}-\frac{n}{a}\int {u^{n-1}e^{au}du} } +C$

$\int {e^{au}\sin \left( {bu} \right) du=\frac{e^{au}}{a^{2}+b^{2}}} \left[ {a\sin \left( {bu} \right)-b\cos \left( {bu} \right)} \right]+C$

$\int {e^{au}\cos \left( {bu} \right) du=\frac{e^{au}}{a^{2}+b^{2}}} \left[ {a\cos \left( {bu} \right)+b\sin \left( {bu} \right)} \right]+C$

$\int {\ln \left( u \right) du=u\ln \left( u \right)-u+C}$

$\int {u^{n}\ln \left( u \right) du=\frac{u^{n+1}}{\left( {n+1} \right)^{2}}\left[ {\left( {n+1} \right)\ln \left( u \right)-1} \right]+C}$

$\int {\frac{1}{u\ln \left( u \right)} du=\ln \left| {\ln \left( u \right)} \right|-u+C}$

$\int {\sinh \left( u \right) du} =\cosh \left( u \right)+C$

$\int {\cosh \left( u \right) du} =\sinh \left( u \right)+C$

$\int {\tanh \left( u \right) du} =\ln \left[ {\cosh \left( u \right)} \right]+C$

$\int {\coth \left( u \right) du} =\ln \left| {\sinh \left( u \right)} \right|+C$

$\int {\mbox{sech}\left( u \right) du} =\tan^{-1}\left| {\sinh \left( u \right)} \right|+C$

$\int {\mbox{csch}\left( u \right) du} =\tan^{-1}\left| {\tanh \left( {\frac{1}{2}u} \right)} \right|+C$

$\int {\mbox{sech}^{\mbox{2}}\left( u \right) du} =\tanh \left( u \right)+C$

$\int {\mbox{c}sch^{\mbox{2}}\left( u \right) du} =-\coth \left( u \right)+C$

$\int {\mbox{sech}\left( u \right)\tanh \left( u \right) du} =-\mbox{sech}\left( u \right)+C$

$\int {\mbox{c}sch\left( u \right)\coth \left( u \right) du} =-\mbox{c}sch\left( u \right)+C$