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Antidifferentiation Summary

$\int u^n\;du=\frac{1}{n+1}\;u^{n+1}+C,\;n \neq -1$

$\int (ax+b)^n\;dx=\frac{1}{a}\;\frac{1}{n+1}(ax+b)^{n+1}+C,\;n \neq -1$

$\int \frac{1}{u}\;du=\ln |u|+C$

$\int \frac{d}{ax+b}\;dx=\frac{d}{a}\ln |ax+b|+C$

$\int \sin u \;du=-\cos u + C$

$\int \sin (ax+b) \;dx=-\frac{1}{a}\cos (ax+b) + C$

$\int \cos u\;du=\sin u + C$

$\int \cos (ax+b)\;dx=\frac{1}{a}\sin (ax+b) + C$

$\int \tan u\;du=\ln |\sec u|+C$

$\int \tan (ax+b)\;dx=\frac{1}{a}\ln |\sec (ax+b)|+C$

$\int \cot u\;du=\ln |\sin u|+C$

$\int \cot (ax+b)\;dx=\frac{1}{a}\ln |\sin (ax+b)|+C$

$\int \sec u\;du=\ln |\sec u +\tan u|+C$

$\int \sec (ax+b)\;dx=\frac{1}{a}\ln |\sec (ax+b) +\tan (ax+b)|+C$

$\int \csc u\;du=\ln |\csc u - \cot u|+C$

$\int \csc (ax+b)\;dx=\frac{1}{a}\ln |\csc (ax+b) - \cot (ax+b)|+C$

$\int \sec^2 u\;du=\tan u +C$

$\int \sec^2 (ax+b)\;dx=\frac{1}{a}\tan (ax+b) +C$

$\int \csc^2 u\;du=-\cot u + C$

$\int \csc^2 (ax+b)\;dx=-\frac{1}{a}\cot (ax+b) + C$

$\int \sec u \tan u\;du=\sec u+C$

$\int \sec (ax+b) \tan (ax+b)\;dx=\frac{1}{a}\sec (ax+b)+C$

$\int \csc u \cot u\;du=-\csc u +C$

$\int \csc (ax+b) \cot (ax+b)\;dx=-\frac{1}{a}\csc (ax+b) +C$

$\int e^u\;du=e^u+C$

$\int e^{ax+b}\;du=\frac{1}{a}e^{ax+b}+C$

$\int a^u\;du=\frac{a^u}{\ln a}+C$

$\int a^{bx+c}\;dx=\frac{a^{bx+c}}{b\ln a}+C$

$\int \frac{1}{\sqrt{a^2-u^2}}\;du=\sin^{-1}\frac{u}{a}+C$

$\int \frac{1}{a^2+u^2}\;du=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$

$\int \frac{1}{u\sqrt{u^2-a^2}}\;du=\frac{1}{a}\sec^{-1}\frac{u}{a}+C$