User:Tarnjp

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\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ constant)}\,\!
\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx
\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx
\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(for } n\neq -1\mbox{)}\,\!
\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 \,{\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
\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
\int \ln {x}\,dx = x \ln {x} - x + C
\int \log_b {x}\,dx = x\log_b {x} - x\log_b {e} + C
\int e^x\,dx = e^x + C
\int a^x\,dx = \frac{a^x}{\ln{a}} + C
\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 \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
\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 \sinh^{-1} x \, dx = x \sinh^{-1} x - \sqrt{x^2+1} + C
\int \cosh^{-1} x \, dx = x \cosh^{-1} x+ \sqrt{x^2-1} + C
\int \tanh^{-1} x \, dx = x \tanh^{-1} x+ \frac{1}{2}\log{(1-x^2)} + C
\int \mbox{csch}^{-1}\,x \, dx = x \mbox{csch}^{-1}\ x+ \log{\left[x\left(\sqrt{1+\frac{1}{x^2}} + 1\right)\right]} + C
\int \mbox{sech}^{-1}\,x \, dx = x \mbox{sech}^{-1}\ x- \tan^{-1}{\left(\frac{x}{x-1}\sqrt{\frac{1-x}{1+x}}\right)} + C
\int \coth^{-1} x \, dx = x \coth^{-1} x+ \frac{1}{2}\log{(x^2-1)} + C
\int_0^\infty{\sqrt{x}\,e^{-x}\,dx} = \frac{1}{2}\sqrt \pi (see also Gamma function)
\int_0^\infty{e^{-x^2}\,dx} = \frac{1}{2}\sqrt \pi (the Gaussian integral)
\int_0^\infty{\frac{x}{e^x-1}\,dx} = \frac{\pi^2}{6} (see also Bernoulli number)
\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 x^{z-1}\,e^{-x}\,dx = \Gamma(z) (where Γ(z) is the Gamma function)
\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right] (where exp[u] is the exponential function eu.)
\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x) (where I0(x) is the modified Bessel function of the first kind)
\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \sqrt{x^2 + y^2}

The method of exhaustion provides a formula for the general case when no antiderivative exists:

\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} )
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