តារាងអាំងតេក្រាលនៃអនុគមន៍អិចស្ប៉ូណង់ស្យែល

ខាងក្រោមនេះគឺជា តារាងអាំងតេក្រាលនៃអនុគមន៍អិចស្ប៉ូណង់ស្យែល​ ៖

សំគាល់ ៖ $x\,$ អាចត្រូវជំនួសដោយ $u\,$ ឬ អថេរផ្សេងទៀត។

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\int {_{\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}} \atop _m} dx= \sum_{n=0}^m\frac{(-1)^n(n+1)^{n-1}}{n!}\Gamma(n+1,- \ln x) + \sum_{n=m+1}^\infty(-1)^na_{mn}\Gamma(n+1,-\ln x) \qquad\mbox{(for }x> 0\mbox{)}$

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

អាំងតេក្រាលកំនត់

$\int_0^1 e^{x\cdot \ln a + (1-x)\cdot \ln b}\;\mathrm{d}x = \int_0^1 \left(\frac{a}{b}\right)^{x}\cdot b\;\mathrm{d}x = \int_0^1 a^{x}\cdot b^{1-x}\;\mathrm{d}x = \frac{a-b}{\ln a - \ln b}$ for $a > 0,\ b > 0,\ a \ne b$ ដែលជាមធ្យមលោការីត

$\int_{0}^{\infty} e^{-ax}\,\mathrm{d}x=\frac{1}{a}$

$\int_{0}^{\infty} e^{-ax^2}\,\mathrm{d}x=\frac{1}{2} \sqrt{\pi \over a} \quad (a>0)$ (អាំងតេក្រាលហ្គោស(Gaussian integral))

$\int_{-\infty}^{\infty} e^{-ax^2}\,\mathrm{d}x=\sqrt{\pi \over a} \quad (a>0)$

$\int_{-\infty}^{\infty} e^{-ax^2} e^{-2bx}\,\mathrm{d}x=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}} \quad (a>0)$

$\int_{-\infty}^{\infty} x e^{-a(x-b)^2}\,\mathrm{d}x= 0$

$\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,\mathrm{d}x=\frac{1}{2} \sqrt{\pi \over a^3} \quad (a>0)$

$\int_{0}^{\infty} x^{n} e^{-ax^2}\,\mathrm{d}x = \begin{cases} \frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\ \frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, a>0) \\ \frac{k!}{2a^{k+1}} & (n=2k+1, a>0) \end{cases}$ $k\,$ ជាចំនួនគត់ (!! ជាហ្វាក់តូរីយែល២ជាន់ (double factorial))

$\int_{0}^{\infty} x^n e^{-ax}\,\mathrm{d}x = \begin{cases} \frac{\Gamma(n+1)}{a^{n+1}} & (n>-1,a>0) \\ \frac{n!}{a^{n+1}} & (n=0,1,2,\ldots,a>0) \\ \end{cases}$

$\int_{0}^{\infty} e^{-ax}\sin bx \, \mathrm{d}x = \frac{b}{a^2+b^2} \quad (a>0)$

$\int_{0}^{\infty} e^{-ax}\cos bx \, \mathrm{d}x = \frac{a}{a^2+b^2} \quad (a>0)$

$\int_{0}^{\infty} xe^{-ax}\sin bx \, \mathrm{d}x = \frac{2ab}{(a^2+b^2)^2} \quad (a>0)$

$\int_{0}^{\infty} xe^{-ax}\cos bx \, \mathrm{d}x = \frac{a^2-b^2}{(a^2+b^2)^2} \quad (a>0)$

$\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x)$ ($I_{0}$ បញ្ជាក់ដោយ អនុគមន៍បេសស៊េល(Bessel function) ប្រភេទទី១)

$\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left( \sqrt{x^2 + y^2} \right)$