La siguiente es una lista de integrales de funciones exponenciales (Agréguese a cada integral una constante arbitraria).

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

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

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

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

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

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

${\displaystyle \int {{e}^{{ax}^{2}\mathrm {+} {bx}\mathrm {+} {c}}}{dx}\mathrm {=} \mathrm {-} {\frac {i}{{e}^{\frac {{b}\mathrm {-} {4}{ac}}{4a}}}}{\sqrt {a}}{\frac {\mathit {\pi }}{2}}{erf}\left({{\frac {{2}{ax}\mathrm {+} {b}}{2{\sqrt {a}}}}i}\right)\;{con}\;\mathrm {\bigtriangleup } {\mathrm {>} }{0}\;{y}\;{a}{\mathrm {>} }{0}}$^[1]​

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

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

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

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

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

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

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

${\displaystyle \int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx={\sqrt {\frac {\pi }{\alpha }}}=I(\alpha )}$

${\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-\alpha x^{2}}dx=(-1)^{n}{\frac {d^{n}}{d\alpha ^{n}}}\ I(\alpha )\ {\xrightarrow {\mbox{si n = 1}}}\ ={\frac {\sqrt {\pi }}{2\alpha ^{3/2}}}}$