daftar integral dari fungsi eksponensial
Video: daftar integral dari fungsi eksponensial
Daftar integral dari fungsi eksponensial GudangMovies21 Rebahinxxi LK21
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
=∫
f
′
(
x
)
e
f
(
x
)
d
x
=
e
f
(
x
)
{\displaystyle \int \mathrm {f} '(x)e^{f(x)}\;\mathrm {d} x=e^{f(x)}}
∫
e
c
x
d
x
=
1
c
e
c
x
{\displaystyle \int e^{cx}\;\mathrm {d} x={\frac {1}{c}}e^{cx}}
∫
a
c
x
d
x
=
1
c
⋅
ln
a
a
c
x
{\displaystyle \int a^{cx}\;\mathrm {d} x={\frac {1}{c\cdot \ln a}}a^{cx}}
for
a
>
0
,
a
≠
1
{\displaystyle a>0,\ a\neq 1}
= Integral melibatkan fungsi eksponensial dan pangkat
=∫
x
e
c
x
d
x
=
e
c
x
c
2
(
c
x
−
1
)
{\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}
∫
x
2
e
c
x
d
x
=
e
c
x
(
x
2
c
−
2
x
c
2
+
2
c
3
)
{\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)}
∫
x
n
e
c
x
d
x
=
1
c
x
n
e
c
x
−
n
c
∫
x
n
−
1
e
c
x
d
x
=
(
∂
∂
c
)
n
e
c
x
c
=
e
c
x
∑
i
=
0
n
(
−
1
)
i
n
!
(
n
−
i
)
!
c
i
+
1
x
n
−
i
{\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}}
∫
e
c
x
x
d
x
=
ln
|
x
|
+
∑
n
=
1
∞
(
c
x
)
n
n
⋅
n
!
{\displaystyle \int {\frac {e^{cx}}{x}}\;\mathrm {d} x=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}}
∫
e
c
x
x
n
d
x
=
1
n
−
1
(
−
e
c
x
x
n
−
1
+
c
∫
e
c
x
x
n
−
1
d
x
)
(for
n
≠
1
)
{\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
=∫
e
c
x
sin
b
x
d
x
=
e
c
x
c
2
+
b
2
(
c
sin
b
x
−
b
cos
b
x
)
=
e
c
x
c
2
+
b
2
sin
(
b
x
−
ϕ
)
cos
(
ϕ
)
=
c
c
2
+
b
2
{\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}}}}}
∫
e
c
x
cos
b
x
d
x
=
e
c
x
c
2
+
b
2
(
c
cos
b
x
+
b
sin
b
x
)
=
e
c
x
c
2
+
b
2
cos
(
b
x
−
ϕ
)
cos
(
ϕ
)
=
c
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}}}}}
∫
e
c
x
sin
n
x
d
x
=
e
c
x
sin
n
−
1
x
c
2
+
n
2
(
c
sin
x
−
n
cos
x
)
+
n
(
n
−
1
)
c
2
+
n
2
∫
e
c
x
sin
n
−
2
x
d
x
{\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}
∫
e
c
x
cos
n
x
d
x
=
e
c
x
cos
n
−
1
x
c
2
+
n
2
(
c
cos
x
+
n
sin
x
)
+
n
(
n
−
1
)
c
2
+
n
2
∫
e
c
x
cos
n
−
2
x
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
=∫
e
c
x
ln
x
d
x
=
1
c
(
e
c
x
ln
|
x
|
−
Ei
(
c
x
)
)
{\displaystyle \int e^{cx}\ln x\;\mathrm {d} x={\frac {1}{c}}\left(e^{cx}\ln |x|-\operatorname {Ei} \,(cx)\right)}
∫
x
e
c
x
2
d
x
=
1
2
c
e
c
x
2
{\displaystyle \int xe^{cx^{2}}\;\mathrm {d} x={\frac {1}{2c}}\;e^{cx^{2}}}
∫
e
−
c
x
2
d
x
=
π
4
c
erf
(
c
x
)
{\displaystyle \int e^{-cx^{2}}\;\mathrm {d} x={\sqrt {\frac {\pi }{4c}}}\operatorname {erf} ({\sqrt {c}}x)}
(
erf
{\displaystyle \operatorname {erf} }
adalah suatu fungsi error)
∫
x
e
−
c
x
2
d
x
=
−
1
2
c
e
−
c
x
2
{\displaystyle \int xe^{-cx^{2}}\;\mathrm {d} x=-{\frac {1}{2c}}e^{-cx^{2}}}
∫
e
−
x
2
x
2
d
x
=
−
e
−
x
2
x
−
π
e
r
f
(
x
)
{\displaystyle \int {\frac {e^{-x^{2}}}{x^{2}}}\;\mathrm {d} x=-{\frac {e^{-x^{2}}}{x}}-{\sqrt {\pi }}\mathrm {erf} (x)}
∫
1
σ
2
π
e
−
1
2
(
x
−
μ
σ
)
2
d
x
=
1
2
(
erf
x
−
μ
σ
2
)
{\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
=∫
e
x
2
d
x
=
e
x
2
(
∑
j
=
0
n
−
1
c
2
j
1
x
2
j
+
1
)
+
(
2
n
−
1
)
c
2
n
−
2
∫
e
x
2
x
2
n
d
x
valid untuk setiap
n
>
0
,
{\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
c
2
j
=
1
⋅
3
⋅
5
⋯
(
2
j
−
1
)
2
j
+
1
=
(
2
j
)
!
j
!
2
2
j
+
1
.
{\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
n
{\displaystyle n}
, karena itu tidak muncul dalam integral.)
∫
x
x
⋅
⋅
x
⏟
m
d
x
=
∑
n
=
0
m
(
−
1
)
n
(
n
+
1
)
n
−
1
n
!
Γ
(
n
+
1
,
−
ln
x
)
+
∑
n
=
m
+
1
∞
(
−
1
)
n
a
m
n
Γ
(
n
+
1
,
−
ln
x
)
(for
x
>
0
)
{\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
a
m
n
=
{
1
jika
n
=
0
,
1
n
!
jika
m
=
1
,
1
n
∑
j
=
1
n
j
a
m
,
n
−
j
a
m
−
1
,
j
−
1
selainnya
{\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
Γ
(
x
,
y
)
{\displaystyle \Gamma (x,y)}
adalah fungsi gamma
∫
1
a
e
λ
x
+
b
d
x
=
x
b
−
1
b
λ
ln
(
a
e
λ
x
+
b
)
{\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
b
≠
0
{\displaystyle b\neq 0}
,
λ
≠
0
{\displaystyle \lambda \neq 0}
, dan
a
e
λ
x
+
b
>
0
.
{\displaystyle ae^{\lambda x}+b>0\,.}
∫
e
2
λ
x
a
e
λ
x
+
b
d
x
=
1
a
2
λ
[
a
e
λ
x
+
b
−
b
ln
(
a
e
λ
x
+
b
)
]
{\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
a
≠
0
{\displaystyle a\neq 0}
,
λ
≠
0
{\displaystyle \lambda \neq 0}
, dan
a
e
λ
x
+
b
>
0
.
{\displaystyle ae^{\lambda x}+b>0\,.}
Integral tertentu
∫
0
1
e
x
⋅
ln
a
+
(
1
−
x
)
⋅
ln
b
d
x
=
∫
0
1
(
a
b
)
x
⋅
b
d
x
=
∫
0
1
a
x
⋅
b
1
−
x
d
x
=
a
−
b
ln
a
−
ln
b
{\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
a
>
0
,
b
>
0
,
a
≠
b
{\displaystyle a>0,\ b>0,\ a\neq b}
, yang merupakan rata-rata logaritme
∫
0
∞
e
a
x
d
x
=
1
−
a
(
Re
(
a
)
<
0
)
{\displaystyle \int _{0}^{\infty }e^{ax}\,\mathrm {d} x={\frac {1}{-a}}\quad (\operatorname {Re} (a)<0)}
∫
0
∞
e
−
a
x
2
d
x
=
1
2
π
a
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)}
(Integral Gaussian)
∫
−
∞
∞
e
−
a
x
2
d
x
=
π
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\sqrt {\pi \over a}}\quad (a>0)}
∫
−
∞
∞
e
−
a
x
2
e
−
2
b
x
d
x
=
π
a
e
b
2
a
(
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)
∫
−
∞
∞
x
e
−
a
(
x
−
b
)
2
d
x
=
b
π
a
(
Re
(
a
)
>
0
)
{\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\frac {\pi }{a}}}\quad (\operatorname {Re} (a)>0)}
∫
−
∞
∞
x
e
−
a
x
2
+
b
x
d
x
=
π
b
2
a
3
/
2
e
b
2
4
a
(
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)}
∫
−
∞
∞
x
2
e
−
a
x
2
d
x
=
1
2
π
a
3
(
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)}
∫
−
∞
∞
x
2
e
−
a
x
2
−
b
x
d
x
=
π
(
2
a
+
b
2
)
4
a
5
/
2
e
b
2
4
a
(
Re
(
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)}
∫
−
∞
∞
x
3
e
−
a
x
2
+
b
x
d
x
=
π
(
6
a
+
b
2
)
b
8
a
7
/
2
e
b
2
4
a
(
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)}
∫
0
∞
x
n
e
−
a
x
2
d
x
=
{
1
2
Γ
(
n
+
1
2
)
/
a
n
+
1
2
(
n
>
−
1
,
a
>
0
)
(
2
k
−
1
)
!
!
2
k
+
1
a
k
π
a
(
n
=
2
k
,
k
integer
,
a
>
0
)
k
!
2
a
k
+
1
(
n
=
2
k
+
1
,
k
integer
,
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)
∫
0
∞
x
n
e
−
a
x
d
x
=
{
Γ
(
n
+
1
)
a
n
+
1
(
n
>
−
1
,
a
>
0
)
n
!
a
n
+
1
(
n
=
0
,
1
,
2
,
…
,
a
>
0
)
{\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}}}
∫
0
1
x
n
e
−
a
x
d
x
=
n
!
a
n
+
1
[
1
−
e
−
a
∑
i
=
0
n
a
i
i
!
]
{\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]}
∫
0
∞
e
−
a
x
b
d
x
=
1
b
a
−
1
b
Γ
(
1
b
)
{\displaystyle \int _{0}^{\infty }e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {1}{b}}}\,\Gamma \left({\frac {1}{b}}\right)}
∫
0
∞
x
n
e
−
a
x
b
d
x
=
1
b
a
−
n
+
1
b
Γ
(
n
+
1
b
)
{\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)}
∫
0
∞
e
−
a
x
sin
b
x
d
x
=
b
a
2
+
b
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,\mathrm {d} x={\frac {b}{a^{2}+b^{2}}}\quad (a>0)}
∫
0
∞
e
−
a
x
cos
b
x
d
x
=
a
a
2
+
b
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,\mathrm {d} x={\frac {a}{a^{2}+b^{2}}}\quad (a>0)}
∫
0
∞
x
e
−
a
x
sin
b
x
d
x
=
2
a
b
(
a
2
+
b
2
)
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }xe^{-ax}\sin bx\,\mathrm {d} x={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
∫
0
∞
x
e
−
a
x
cos
b
x
d
x
=
a
2
−
b
2
(
a
2
+
b
2
)
2
(
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)}
∫
0
2
π
e
x
cos
θ
d
θ
=
2
π
I
0
(
x
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}
(
I
0
{\displaystyle I_{0}}
adalah modifikasi fungsi Bessel dari jenis pertama)
∫
0
2
π
e
x
cos
θ
+
y
sin
θ
d
θ
=
2
π
I
0
(
x
2
+
y
2
)
{\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
Wolfram Mathematica Online Integrator
V. H. Moll, The Integrals in Gradshteyn and Ryzhik