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
sunting
∫
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}}
a
>
0
,
a
≠
1
{\displaystyle a>0,\ a\neq 1}
Integral melibatkan fungsi eksponensial dan pangkat
sunting
∫
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
sunting
∫
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
sunting
∫
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} }
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)}
∫
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}
∫
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)}
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)\,}
b
≠
0
{\displaystyle b\neq 0}
λ
≠
0
{\displaystyle \lambda \neq 0}
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]\,}
a
≠
0
{\displaystyle a\neq 0}
λ
≠
0
{\displaystyle \lambda \neq 0}
a
e
λ
x
+
b
>
0
.
{\displaystyle ae^{\lambda x}+b>0\,.}
∫
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}}}
a
>
0
,
b
>
0
,
a
≠
b
{\displaystyle a>0,\ b>0,\ a\neq b}
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)}
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}}}
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}}
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)}
![{\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]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/278961ab24efa75c218ed9fc570b0f9a7dac350f)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8646de30e21caf89876f564001c275877a219bf7)
