\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \neq 0 \mbox{, constant)}\,\!
\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx
\int f'(x)g(x)\,dx = f(x)g(x) - \int f(x)g'(x)\,dx
\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C
\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C
\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(for } n\neq -1\mbox{)}\,\!

Rational functions

\int \,{\rm d}x = x + C
\int x^n\,{\rm d}x = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ if }n \ne -1
\int {dx \over x} = \ln{\left|x\right|} + C
\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C

Irrational functions

\int {dx \over \sqrt{a^2-x^2}} = \sin^{-1} {x \over a} + C
\int {-dx \over \sqrt{a^2-x^2}} = \cos^{-1} {x \over a} + C
\int {dx \over x \sqrt{x^2-a^2}} = {1 \over a} \sec^{-1} {|x| \over a} + C

Logarithms

\int \ln {x}\,dx = x \ln {x} - x + C
\int \log_b {x}\,dx = x\log_b {x} - x\log_b {e} + C

Exponential functions

\int e^x\,dx = e^x + C
\int a^x\,dx = \frac{a^x}{\ln{a}} + C

Trigonometric functions

\int \sin{x}\, dx = -\cos{x} + C
\int \cos{x}\, dx = \sin{x} + C
\int \tan{x} \, dx = -\ln{\left| \cos {x} \right|} + C
\int \cot{x} \, dx = \ln{\left| \sin{x} \right|} + C
\int \sec{x} \, dx = \ln{\left| \sec{x} + \tan{x}\right|} + C
\int \csc{x} \, dx = \ln{\left| \csc{x} - \cot{x}\right|} + C
\int \sec^2 x \, dx = \tan x + C
\int \csc^2 x \, dx = -\cot x + C
\int \sec{x} \, \tan{x} \, dx = \sec{x} + C
\int \csc{x} \, \cot{x} \, dx = - \csc{x} + C
\int \sin^2 x \, dx = \frac{1}{2}(x - \sin x \cos x) + C
\int \cos^2 x \, dx = \frac{1}{2}(x + \sin x \cos x) + C
\int \sec^3 x \, dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C
\int \sin^n x \, dx = - \frac{\sin^{n-1} {x} \cos {x}}{n} + \frac{n-1}{n} \int \sin^{n-2}{x} \, dx
\int \cos^n x \, dx = \frac{\cos^{n-1} {x} \sin {x}}{n} + \frac{n-1}{n} \int \cos^{n-2}{x} \, dx
\int \arctan{x} \, dx = x \, \arctan{x} - \frac{1}{2} \ln{\left| 1 + x^2\right|} + C

Hyperbolic functions

\int \sinh x \, dx = \cosh x + C
\int \cosh x \, dx = \sinh x + C
\int \tanh x \, dx = \ln| \cosh x | + C
\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C
\int \mbox{sech}\,x \, dx = \arctan(\sinh x) + C
\int \coth x \, dx = \ln| \sinh x | + C
\int \mbox{sech}^2 x\, dx = \tanh x + C

Inverse hyperbolic functions

\int \operatorname{arcsinh} x \, dx = x \operatorname{arcsinh} x - \sqrt{x^2+1} + C
\int \operatorname{arccosh} x \, dx = x \operatorname{arccosh} x - \sqrt{x^2-1} + C
\int \operatorname{arctanh} x \, dx = x \operatorname{arctanh} x + \frac{1}{2}\log{(1-x^2)} + C
\int \operatorname{arccsch}\,x \, dx = x \operatorname{arccsch} x+ \log{\left[x\left(\sqrt{1+\frac{1}{x^2}} + 1\right)\right]} + C
\int \operatorname{arcsech}\,x \, dx = x \operatorname{arcsech} x- \arctan{\left(\frac{x}{x-1}\sqrt{\frac{1-x}{1+x}}\right)} + C
\int \operatorname{arccoth}\,x \, dx = x \operatorname{arccoth} x+ \frac{1}{2}\log{(x^2-1)} + C

Definite integrals lacking closed-form antiderivatives

\int_0^\infty{\sqrt{x}\,e^{-x}\,dx} = \frac{1}{2}\sqrt \pi 
\int_0^\infty{e^{-x^2}\,dx} = \frac{1}{2}\sqrt \pi 
\int_0^\infty{\frac{x}{e^x-1}\,dx} = \frac{\pi^2}{6} 
\int_0^\infty{\frac{x^3}{e^x-1}\,dx} = \frac{\pi^4}{15}
\int_0^\infty\frac{\sin(x)}{x}\,dx=\frac{\pi}{2}
\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot n}\frac{\pi}{2} (if n is an even integer and \scriptstyle{n \ge 2})
\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (n-1)}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot n} (if \scriptstyle{n} is an odd integer and \scriptstyle{n \ge 3} )
\int_0^\infty\frac{\sin^2{x}}{x^2}\,dx=\frac{\pi}{2}
\int_0^\infty x^{z-1}\,e^{-x}\,dx = \Gamma(z) 
\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right]
\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x) 
\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left(\sqrt{x^2 + y^2}\right)
\int_{-\infty}^{\infty}{(1 + x^2/\nu)^{-(\nu + 1)/2}dx} = \frac { \sqrt{\nu \pi} \ \Gamma(\nu/2)} {\Gamma((\nu + 1)/2))}\, (\nu > 0\,,
\int_a^b{f(x)\,dx} = (b - a) \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^{2^n - 1} {\left( { - 1} \right)^{m + 1} } } 2^{ - n} f(a + m\left( {b - a} \right)2^{-n} )

\begin{align} \int_0^1 x^{-x}\,dx &= \sum_{n=1}^\infty n^{-n} &&(= 1.291285997\dots)\\ \int_0^1 x^x \,dx &= \sum_{n=1}^\infty -(-1)^nn^{-n} &&(= 0.783430510712\dots) \end{align}