Daftar integral dari fungsi eksponensial

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

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

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

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

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

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

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

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

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

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

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

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

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

$\int e^{-c x^2 }\; dx= \sqrt{\frac{\pi}{4c}} \mbox{erf}(\sqrt{c} x)$ ( $\mbox{erf}$ adalah fungsi kesalahan/error function)

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

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

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

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

Integral tertentu

$\int_{0}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0)$ (integral Gauss)

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

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

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

$\int_{0}^{\infty} x^{n} e^{-ax^2}\,dx = \begin{cases} \frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\ \frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{bilangan bulat}, a>0) \\ \frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{bilangan bulat}, a>0) \end{cases}$

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