Primitivele funcțiilor exponențiale

Următorul articol este o listă de integrale (primitive) de funcții exponențiale. Pentru o listă cu mai multe integrale, vezi tabel de integrale și lista integralelor.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\sqrt {\pi \over a}}}$ (Integrala gaussiană)

${\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,dx=b{\sqrt {\pi \over a}}}$

${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}}$

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

${\displaystyle \int _{-\infty }^{\infty }e^{-{x^{2}}/{a^{2}}}\cos bx\,dx=a{\sqrt {\pi }}(\sin {a^{2}b^{2} \over 4}+\cos {a^{2}b^{2} \over 4})}$

${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}$ (${\displaystyle I_{0}}$ este funcția Bessel de speța I modificată)

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

${\displaystyle \int _{0}^{\infty }x^{a}e^{-bx}dx={\frac {a!}{b^{a+1}}}}$