**Daftar integral (antiderivatif) dari fungsi eksponensial**. Untuk daftar lengkap fungsi integral, lihat Tabel integral.

Dalam semua rumus, konstanta *a* diasumsikan bukan nol.

Integral tak tentu
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Integral tak tentu adalah fungsi-fungsi antiderivatif. Sebuah konstanta (yaitu konstanta integrasi) dapat ditambahkan pada sisi kanan dari rumus ini, tetapi tidak dituliskan di sini demi kesederhanaan.

### Integral melibatkan hanya fungsi eksponensial

{\displaystyle \int \mathrm {f} '(x)e^{f(x)}\;\mathrm {d} x=e^{f(x)}}
{\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}} for {\displaystyle a>0,\ a\neq 1}
### Integral melibatkan fungsi eksponensial dan pangkat

{\displaystyle \int xe^{cx}\;\mathrm {d} x={\frac {e^{cx}}{c^{2}}}(cx-1)}
\int xe^{-cx}\; \mathrm{d}x =x \frac{1}{-c}e^{-cx}
{\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=\left({\frac {\partial }{\partial c}}\right)^{n}{\frac {e^{cx}}{c}}=e^{cx}\sum \_{i=0}^{n}(-1)^{i}\,{\frac {n!}{(n-i)!\,c^{i+1}}}\,x^{n-i}}
{\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)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
### Integral melibatkan fungsi eksponensial dan trigonometri

{\displaystyle \int e^{cx}\sin bx\;\mathrm {d} x={\frac {e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)={\frac {e^{cx}}{\sqrt {c^{2}+b^{2}}}}\sin(bx-\phi )\qquad \cos(\phi )={\frac {c}{\sqrt {c^{2}+b^{2}}}}}
{\displaystyle \int e^{cx}\cos bx\;\mathrm {d} x={\frac {e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)={\frac {e^{cx}}{\sqrt {c^{2}+b^{2}}}}\cos(bx-\phi )\qquad \cos(\phi )={\frac {c}{\sqrt {c^{2}+b^{2}}}}}
{\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}
### Integral melibatkan fungsi kesalahan

{\displaystyle \int e^{cx}\ln x\;\mathrm {d} x={\frac {1}{c}}\left(e^{cx}\ln |x|-\operatorname {Ei} \,(cx)\right)}
{\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}}}\operatorname {erf} ({\sqrt {c}}x)} ({\displaystyle \operatorname {erf} } adalah suatu fungsi error)
{\displaystyle \int xe^{-cx^{2}}\;\mathrm {d} x=-{\frac {1}{2c}}e^{-cx^{2}}}
{\displaystyle \int {\frac {e^{-x^{2}}}{x^{2}}}\;\mathrm {d} x=-{\frac {e^{-x^{2}}}{x}}-{\sqrt {\pi }}\mathrm {erf} (x)}
{\displaystyle \int {{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}}\;\mathrm {d} x={\frac {1}{2}}\left(\operatorname {erf} \,{\frac {x-\mu }{\sigma {\sqrt {2}}}}\right)}
### Integral lain-lain

{\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\quad {\mbox{valid untuk setiap }}n>0,}
di mana {\displaystyle c\_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)\,!}{j!\,2^{2j+1}}}\ .}
(Perhatikan bahwa nilai ekspresi ini *independen* atau tidak tergantung dari nilai {\displaystyle n}, karena itu tidak muncul dalam integral.)
{\displaystyle {\int \underbrace {x^{x^{\cdot ^{\cdot ^{x}}}}} \_{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{)}}}}
di mana {\displaystyle a\_{mn}={\begin{cases}1&{\text{jika }}n=0,\\{\frac {1}{n!}}&{\text{jika }}m=1,\\{\frac {1}{n}}\sum \_{j=1}^{n}ja\_{m,n-j}a\_{m-1,j-1}&{\text{selainnya}}\end{cases}}}
dan {\displaystyle \Gamma (x,y)} adalah fungsi gamma
{\displaystyle \int {\frac {1}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)\,} ketika {\displaystyle b\neq 0}, {\displaystyle \lambda \neq 0}, dan {\displaystyle ae^{\lambda x}+b>0\,.}
{\displaystyle \int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]\,} ketika {\displaystyle a\neq 0}, {\displaystyle \lambda \neq 0}, dan {\displaystyle ae^{\lambda x}+b>0\,.}

Integral tertentu
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{\displaystyle \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}}} untuk {\displaystyle a>0,\ b>0,\ a\neq b}, yang merupakan rata-rata logaritme
{\displaystyle \int \_{0}^{\infty }e^{ax}\,\mathrm {d} x={\frac {1}{-a}}\quad (\operatorname {Re} (a)<0)}
{\displaystyle \int \_{0}^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)} (Integral Gaussian)
{\displaystyle \int \_{-\infty }^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\sqrt {\pi \over a}}\quad (a>0)}
{\displaystyle \int \_{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,\mathrm {d} x={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)} (lihat Integral suatu fungsi Gaussian)
{\displaystyle \int \_{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\frac {\pi }{a}}}\quad (\operatorname {Re} (a)>0)}
{\displaystyle \int \_{-\infty }^{\infty }xe^{-ax^{2}+bx}\,\mathrm {d} x={\frac {{\sqrt {\pi }}b}{2a^{3/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}
{\displaystyle \int \_{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (a>0)}
{\displaystyle \int \_{-\infty }^{\infty }x^{2}e^{-ax^{2}-bx}\,\mathrm {d} x={\frac {{\sqrt {\pi }}(2a+b^{2})}{4a^{5/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}
{\displaystyle \int \_{-\infty }^{\infty }x^{3}e^{-ax^{2}+bx}\,\mathrm {d} x={\frac {{\sqrt {\pi }}(6a+b^{2})b}{8a^{7/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}
{\displaystyle \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,k\;{\text{integer}},a>0)\\{\frac {k!}{2a^{k+1}}}&(n=2k+1,k\;{\text{integer}},a>0)\end{cases}}} (!! merupakan faktorial ganda)
{\displaystyle \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}}}
{\displaystyle \int \_{0}^{1}x^{n}e^{-ax}\,\mathrm {d} x={\frac {n!}{a^{n+1}}}\left[1-e^{-a}\sum \_{i=0}^{n}{\frac {a^{i}}{i!}}\right]}
{\displaystyle \int \_{0}^{\infty }e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {1}{b}}}\,\Gamma \left({\frac {1}{b}}\right)}
{\displaystyle \int \_{0}^{\infty }x^{n}e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {n+1}{b}}}\,\Gamma \left({\frac {n+1}{b}}\right)}
{\displaystyle \int \_{0}^{\infty }e^{-ax}\sin bx\,\mathrm {d} x={\frac {b}{a^{2}+b^{2}}}\quad (a>0)}
{\displaystyle \int \_{0}^{\infty }e^{-ax}\cos bx\,\mathrm {d} x={\frac {a}{a^{2}+b^{2}}}\quad (a>0)}
{\displaystyle \int \_{0}^{\infty }xe^{-ax}\sin bx\,\mathrm {d} x={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
{\displaystyle \int \_{0}^{\infty }xe^{-ax}\cos bx\,\mathrm {d} x={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
{\displaystyle \int \_{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I\_{0}(x)} ({\displaystyle I\_{0}} adalah modifikasi fungsi Bessel dari jenis pertama)
{\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)}

Pranala luar
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* Wolfram Mathematica Online Integrator
* V. H. Moll, The Integrals in Gradshteyn and Ryzhik

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