قائمة بتكاملات التوابع الأسية

هذه قائمة بتكاملات الدوال الأسية:

${\displaystyle{\displaystyle\int e^{cx}\;dx={\frac{1}{c}}e^{cx}}}$

${\displaystyle{\displaystyle\int a^{cx}\;dx={\frac{1}{c\ln a}}a^{cx}\qquad{\mbox{(for }}a>0,{\mbox{ }}a\neq 1{\mbox{)}}}}$

${\displaystyle{\displaystyle\int xe^{cx}\;dx={\frac{e^{cx}}{c^{2}}}(cx-1)}}$

${\displaystyle{\displaystyle\int x^{2}e^{cx}\;dx=e^{cx}\left({\frac{x^{2}}{c}}-{\frac{2x}{c^{2}}}+{\frac{2}{c^{3}}}\right)}}$

${\displaystyle{\displaystyle\int x^{n}e^{cx}\;dx={\frac{1}{c}}x^{n}e^{cx}-{\frac{n}{c}}\int x^{n-1}e^{cx}dx}}$

${\displaystyle{\displaystyle\int{\frac{e^{cx}\;dx}{x}}=\ln|x|+\sum_{i=1}^{\infty}{\frac{(cx)^{i}}{i\cdot i!}}}}$

${\displaystyle{\displaystyle\int{\frac{e^{cx}\;dx}{x^{n}}}={\frac{1}{n-1}}\left(-{\frac{e^{cx}}{x^{n-1}}}+c\int{\frac{e^{cx}}{x^{n-1}}}\,dx\right)\qquad{\mbox{(for }}n\neq 1{\mbox{)}}}}$

${\displaystyle{\displaystyle\int e^{cx}\ln x\;dx={\frac{1}{c}}e^{cx}\ln|x|-\operatorname{Ei}\,(cx)}}$

${\displaystyle{\displaystyle\int e^{cx}\sin bx\;dx={\frac{e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)}}$

${\displaystyle{\displaystyle\int e^{cx}\cos bx\;dx={\frac{e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)}}$

${\displaystyle{\displaystyle\int e^{cx}\sin^{n}x\;dx={\frac{e^{cx}\sin^{n-1}x}{c^{2}+n^{2}}}(c\sin x-n\cos x)+{\frac{n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\sin^{n-2}x\;dx}}$

${\displaystyle{\displaystyle\int e^{cx}\cos^{n}x\;dx={\frac{e^{cx}\cos^{n-1}x}{c^{2}+n^{2}}}(c\cos x+n\sin x)+{\frac{n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\cos^{n-2}x\;dx}}$

${\displaystyle{\displaystyle\int xe^{cx^{2}}\;dx={\frac{1}{2c}}\;e^{cx^{2}}}}$

${\displaystyle{\displaystyle\int{1\over\sigma{\sqrt{2\pi}}}\,e^{-{(x-\mu)^{2}/2\sigma^{2}}}\;dx={\frac{1}{2\sigma}}(1+{\mbox{erf}}\,{\frac{x-\mu}{\sigma{\sqrt{2}}}})}}$

${\displaystyle{\displaystyle\int e^{x^{2}}\,dx=e^{x^{2}}\left(\sum_{j=0}^{n-1}c_{2j}\,{\frac{1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int{\frac{e^{x^{2}}}{x^{2n}}}\;dx\quad{\mbox{valid for }}n>0,}}$

where ${\displaystyle{\displaystyle c_{2j}={\frac{1\cdot 3\cdot 5\cdots(2j-1)}{2^{j+1}}}={\frac{2j\,!}{j!\,2^{2j+1}}}\ .}}$

${\displaystyle{\displaystyle\int_{-\infty}^{\infty}e^{-ax^{2}}\,dx={\sqrt{\pi\over a}}}}$

${\displaystyle{\displaystyle\int_{0}^{\infty}x^{2n}e^{-{x^{2}}/{a^{2}}}\,dx={\sqrt{\pi}}{(2n)!\over{n!}}{\left({\frac{a}{2}}\right)}^{2n+1}}}$