តារាងអាំងតេក្រាលនៃអនុគមន៍អិចស្ប៉ូណង់ស្យែល

ខាងក្រោមនេះគឺជា តារាងអាំងតេក្រាលនៃអនុគមន៍អិចស្ប៉ូណង់ស្យែល​ ៖

សំគាល់ ៖ ${\displaystyle x\,}$ អាចត្រូវជំនួសដោយ ${\displaystyle u\,}$ ឬ អថេរផ្សេងទៀត។

${\displaystyle \int e^{cx}\;\mathrm {d} x={\frac {1}{c}}e^{cx}}$

${\displaystyle \int a^{cx}\;\mathrm {d} x={\frac {1}{c\cdot \ln a}}a^{cx}}$ ចំពោះ ${\displaystyle a>0,\ a\neq 1}$

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

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

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

${\displaystyle \int {\frac {e^{cx}}{x}}\;\mathrm {d} x=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}}$

${\displaystyle \int {\frac {e^{cx}}{x^{n}}}\;\mathrm {d} x={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,\mathrm {d} x\right)\,}$ ចំពោះ ${\displaystyle n\neq 1\,}$

${\displaystyle \int e^{cx}\ln x\;\mathrm {d} x={\frac {1}{c}}e^{cx}\ln |x|-\operatorname {Ei} \,(cx)}$ ដែល ${\displaystyle \operatorname {Ei} \,(\cdot )}$ គឺជាអនុគមន៍អាំងតេក្រាលអិចស្ប៉ូណង់ស្យែល

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

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

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

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

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

${\displaystyle \int e^{-cx^{2}}\;\mathrm {d} x={\sqrt {\frac {\pi }{4c}}}{\mbox{erf}}({\sqrt {c}}x)}$ ដែល ${\displaystyle {\mbox{erf}}(\cdot )}$ ជាអនុគមន៍កំហុស(error function)

${\displaystyle \int xe^{-cx^{2}}\;\mathrm {d} x=-{\frac {1}{2c}}e^{-cx^{2}}}$

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

${\displaystyle \int e^{x^{2}}\,\mathrm {d} x=e^{x^{2}}\left(\sum _{j=0}^{n-1}c_{2j}\,{\frac {1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int {\frac {e^{x^{2}}}{x^{2n}}}\;\mathrm {d} x\,}$ ចំពោះ ${\displaystyle n>0\,}$

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

${\displaystyle \int {_{\underbrace {x^{x^{\cdot ^{\cdot ^{x}}}}} } \atop _{m}}dx=\sum _{n=0}^{m}{\frac {(-1)^{n}(n+1)^{n-1}}{n!}}\Gamma (n+1,-\ln x)+\sum _{n=m+1}^{\infty }(-1)^{n}a_{mn}\Gamma (n+1,-\ln x)\qquad {\mbox{(for }}x>0{\mbox{)}}}$ ^[១]

ដែល ${\displaystyle a_{mn}={\begin{cases}1&{\text{if }}n=0,\\{\frac {1}{n!}}&{\text{if }}m=1,\\{\frac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}}$