Daftar integral dari fungsi eksponensial

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

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

$\int {\mathrm{f}}^{\prime }(x)e^{f(x)}\mathrm{d}x=e^{f(x)}$

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

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

Integral melibatkan fungsi eksponensial dan pangkat

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

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

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

$\int x^n{e}^{cx}\mathrm{d}x=\frac{1}{c}{x}^n{e}^{cx}-\frac{n}{c}\int x^{n-1}{e}^{cx}\mathrm{d}x={\left(\frac{\partial }{\partial c}\right)}^n\frac{e^{cx}}{c}=e^{cx}\sum _{i=0}^n(-1{)}^i\frac{n!}{(n-i)!c^{i+1}}{x}^{n-i}$

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

$\int \frac{e^{cx}}{x^n}\mathrm{d}x=\frac{1}{n-1}(-\frac{e^{cx}}{x^{n-1}}+c\int \frac{e^{cx}}{x^{n-1}}\mathrm{d}x)\text{(for }n\ne 1\text{)}$

Integral melibatkan fungsi eksponensial dan trigonometri

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

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

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

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

Integral melibatkan fungsi kesalahan

$\int e^{cx}\ln x\mathrm{d}x=\frac{1}{c}(e^{cx}\ln |x|-\mathrm{Ei}(cx))$

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

$\int e^{-c{x}^2}\mathrm{d}x=\sqrt{\frac{\pi }{4c}}\mathrm{erf}(\sqrt{c}x)$ ($\mathrm{erf}$ adalah suatu fungsi error)

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

$\int \frac{e^{-x^2}}{x^2}\mathrm{d}x=-\frac{e^{-x^2}}{x}-\sqrt{\pi }\mathrm{e}\mathrm{r}\mathrm{f}(x)$

$\int \frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}\mathrm{d}x=\frac{1}{2}(\mathrm{erf}\frac{x-\mu }{\sigma \sqrt{2}})$

Integral lain-lain

$\int e^{x^2}\mathrm{d}x=e^{x^2}\left(\sum _{j=0}^{n-1}{c}_{2j}\frac{1}{x^{2j+1}}\right)+(2n-1)c_{2n-2}\int \frac{e^{x^2}}{x^{2n}}\mathrm{d}x\text{valid untuk setiap }n>0,$

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

(Perhatikan bahwa nilai ekspresi ini independen atau tidak tergantung dari nilai $n$, karena itu tidak muncul dalam integral.)

$\int \underset{m}{\underbrace{x^{x^{{\cdot }^{{\cdot }^x}}}}}dx=\sum _{n=0}^m\frac{(-1{)}^n(n+1{)}^{n-1}}{n!}\mathrm{\Gamma }(n+1,-\ln x)+\sum _{n=m+1}^{\mathrm{\infty }}(-1{)}^n{a}_{mn}\mathrm{\Gamma }(n+1,-\ln x)\text{(for }x>0\text{)}$

di mana $a_{mn}=\{\begin{array}{cc}1& \text{jika }n=0,\\ \frac{1}{n!}& \text{jika }m=1,\\ \frac{1}{n}\sum _{j=1}^{n}j{a}_{m,n-j}{a}_{m-1,j-1}& \text{selainnya}\end{array}$

dan $\mathrm{\Gamma }(x,y)$ adalah fungsi gamma

$\int \frac{1}{a{e}^{\lambda x}+b}\mathrm{d}x=\frac{x}{b}-\frac{1}{b\lambda }\ln (a{e}^{\lambda x}+b)$ ketika $b\ne 0$, $\lambda \ne 0$, dan $a{e}^{\lambda x}+b>0.$

$\int \frac{e^{2\lambda x}}{a{e}^{\lambda x}+b}\mathrm{d}x=\frac{1}{a^2\lambda }[a{e}^{\lambda x}+b-b\ln (a{e}^{\lambda x}+b)]$ ketika $a\ne 0$, $\lambda \ne 0$, dan $a{e}^{\lambda x}+b>0.$

Integral tertentu

$\underset{0}{\overset{1}{\int }}{e}^{x\cdot \ln a+(1-x)\cdot \ln b}\mathrm{d}x=\underset{0}{\overset{1}{\int }}{\left(\frac{a}{b}\right)}^x\cdot b\mathrm{d}x=\underset{0}{\overset{1}{\int }}{a}^x\cdot b^{1-x}\mathrm{d}x=\frac{a-b}{\ln a-\ln b}$ untuk $a>0,b>0,a\ne b$, yang merupakan rata-rata logaritme

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{ax}\mathrm{d}x=\frac{1}{-a}(\mathrm{Re}(a)<0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}\mathrm{d}x=\frac{1}{2}\sqrt{\frac{\pi }{a}}(a>0)$ (Integral Gaussian)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}\mathrm{d}x=\sqrt{\frac{\pi }{a}}(a>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}{e}^{-2bx}\mathrm{d}x=\sqrt{\frac{\pi }{a}}{e}^{\frac{b^2}{a}}(a>0)$ (lihat Integral suatu fungsi Gaussian)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{-a(x-b{)}^2}\mathrm{d}x=b\sqrt{\frac{\pi }{a}}(\mathrm{Re}(a)>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{-a{x}^2+bx}\mathrm{d}x=\frac{\sqrt{\pi }b}{2a^{3/2}}{e}^{\frac{b^2}{4a}}(\mathrm{Re}(a)>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-a{x}^2}\mathrm{d}x=\frac{1}{2}\sqrt{\frac{\pi }{a^3}}(a>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-a{x}^2-bx}\mathrm{d}x=\frac{\sqrt{\pi }(2a+b^2)}{4a^{5/2}}{e}^{\frac{b^2}{4a}}(\mathrm{Re}(a)>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^3{e}^{-a{x}^2+bx}\mathrm{d}x=\frac{\sqrt{\pi }(6a+b^2)b}{8a^{7/2}}{e}^{\frac{b^2}{4a}}(\mathrm{Re}(a)>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-a{x}^2}\mathrm{d}x=\{\begin{array}{cc}\frac{1}{2}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}}& (n>-1,a>0)\\ \frac{(2k-1)!!}{2^{k+1}{a}^k}\sqrt{\frac{\pi }{a}}& (n=2k,k\text{integer},a>0)\\ \frac{k!}{2a^{k+1}}& (n=2k+1,k\text{integer},a>0)\end{array}$ (!! merupakan faktorial ganda)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-ax}\mathrm{d}x=\{\begin{array}{cc}\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}& (n>-1,a>0)\\ \frac{n!}{a^{n+1}}& (n=0,1,2,\dots ,a>0)\end{array}$

$\underset{0}{\overset{1}{\int }}{x}^n{e}^{-ax}\mathrm{d}x=\frac{n!}{a^{n+1}}[1-e^{-a}\sum _{i=0}^n\frac{a^i}{i!}]$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^b}dx=\frac{1}{b}{a}^{-\frac{1}{b}}\mathrm{\Gamma }\left(\frac{1}{b}\right)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-a{x}^b}dx=\frac{1}{b}{a}^{-\frac{n+1}{b}}\mathrm{\Gamma }\left(\frac{n+1}{b}\right)$

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

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}x{e}^{-ax}\sin bx\mathrm{d}x=\frac{2ab}{(a^2+b^2{)}^2}(a>0)$

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

$\underset{0}{\overset{2\pi }{\int }}{e}^{x\cos \theta }d\theta =2\pi I_0(x)$ ($I_0$ adalah modifikasi fungsi Bessel dari jenis pertama)

$\underset{0}{\overset{2\pi }{\int }}{e}^{x\cos \theta +y\sin \theta }d\theta =2\pi I_0\left(\sqrt{x^2+y^2}\right)$

Pranala luar

• Wolfram Mathematica Online Integrator
• V. H. Moll, The Integrals in Gradshteyn and Ryzhik