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Matematika Wiki
974
pages
Explore
Main Page
All Pages
Community
Interactive Maps
Recent Blog Posts
Wiki Content
Recently Changed Pages
Primjena L - funkcija u teoriji brojeva/2
Broj 100
Zadatak 244
Primjena L - funkcija u teoriji brojeva/1
Deltoid
Primjena L - funkcija u teoriji brojeva
Starogrčka matematika/ 5
Algebra
Zadatak 6
Logaritamske jednačine
Zadatak 15
Zbir kvadrata prvih n prirodnih brojeva
Logaritanska funkcija
Zadatak 76
Zadatak 77
Zadatak
Zadatak 6
Zadatak 7
Zadatak 13
Zadatak 15
Zadatak 16
Zadatak 19
Zadatak 21
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in:
Integrali
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Matematička analiza
Popis integrala eksponencijalnih funkcija
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∫
e
c
x
d
x
=
1
c
e
c
x
{\displaystyle \int e^{cx}\;dx = \frac{1}{c} e^{cx}}
, ali
∫
e
2
x
d
x
=
2
e
2
x
{\displaystyle \int e^{2x}\;dx=2e^{2x}}
∫
a
c
x
d
x
=
1
c
ln
a
a
c
x
(za
a
>
0
,
a
≠
1
)
{\displaystyle \int a^{cx}\;dx={\frac {1}{c\ln a}}a^{cx}\qquad {\mbox{(za }}a>0,{\mbox{ }}a\neq 1{\mbox{)}}}
∫
x
e
c
x
d
x
=
e
c
x
c
2
(
c
x
−
1
)
{\displaystyle \int xe^{cx}\;dx = \frac{e^{cx}}{c^2}(cx-1)}
∫
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}\;dx = 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
{\displaystyle \int x^n e^{cx}\; dx = \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} dx}
∫
e
c
x
x
d
x
=
ln
|
x
|
+
∑
i
=
1
∞
(
c
x
)
i
i
⋅
i
!
{\displaystyle \int {\frac {e^{cx}}{x}}\;dx=\ln |x|+\sum _{i=1}^{\infty }{\frac {(cx)^{i}}{i\cdot i!}}}
∫
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
)
(za
n
≠
1
)
{\displaystyle \int {\frac {e^{cx}}{x^{n}}}\;dx={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,dx\right)\qquad {\mbox{(za }}n\neq 1{\mbox{)}}}
∫
e
c
x
ln
x
d
x
=
1
c
e
c
x
ln
|
x
|
−
Ei
(
c
x
)
{\displaystyle \int e^{cx}\ln x\; dx = \frac{1}{c}e^{cx}\ln|x|-\operatorname{Ei}\,(cx)}
∫
e
c
x
sin
b
x
d
x
=
e
c
x
c
2
+
b
2
(
c
sin
b
x
−
b
cos
b
x
)
{\displaystyle \int e^{cx}\sin bx\; dx = \frac{e^{cx}}{c^2+b^2}(c\sin bx - b\cos bx)}
∫
e
c
x
cos
b
x
d
x
=
e
c
x
c
2
+
b
2
(
c
cos
b
x
+
b
sin
b
x
)
{\displaystyle \int e^{cx}\cos bx\; dx = \frac{e^{cx}}{c^2+b^2}(c\cos bx + b\sin bx)}
∫
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\; 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}
∫
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\; 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}
∫
x
e
c
x
2
d
x
=
1
2
c
e
c
x
2
{\displaystyle \int x e^{c x^2 }\; dx= \frac{1}{2c} \; e^{c x^2}}
∫
e
−
c
x
2
d
x
=
π
4
c
erf
(
c
x
)
{\displaystyle \int e^{-cx^{2}}\;dx={\sqrt {\frac {\pi }{4c}}}{\mbox{erf}}({\sqrt {c}}x)}
(
erf
{\displaystyle {\mbox{erf}}}
je
funkcija greške
(
error function
))
∫
x
e
−
c
x
2
d
x
=
−
1
2
c
e
−
c
x
2
{\displaystyle \int xe^{-cx^{2}}\;dx=-{\frac {1}{2c}}e^{-cx^{2}}}
∫
1
σ
2
π
e
−
(
x
−
μ
)
2
/
2
σ
2
d
x
=
1
2
(
1
+
erf
x
−
μ
σ
2
)
{\displaystyle \int {1 \over \sigma {\sqrt {2\pi }}}\,e^{-{(x-\mu )^{2}/2\sigma ^{2}}}\;dx={\frac {1}{2}}(1+{\mbox{erf}}\,{\frac {x-\mu }{\sigma {\sqrt {2}}}})}
∫
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
valjano za
n
>
0
,
{\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{valjano za }}n>0,}
pri čemu je
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}}}\ .}
Određeni integrali
[
]
∫
0
∞
e
−
a
x
2
d
x
=
1
2
π
a
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)}
(
Gaussov integral
)
∫
−
∞
∞
e
−
a
x
2
d
x
=
π
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\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}\,dx={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)}
∫
−
∞
∞
x
e
−
a
(
x
−
b
)
2
d
x
=
b
π
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,dx=b{\sqrt {\pi \over a}}\quad (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}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (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
cijeli broj
,
a
>
0
)
k
!
2
a
k
+
1
(
n
=
2
k
+
1
,
k
cijeli broj
,
a
>
0
)
{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2}}\,dx={\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{cijeli broj}},a>0)\\{\frac {k!}{2a^{k+1}}}&(n=2k+1,k\;{\text{cijeli broj}},a>0)\end{cases}}}
(!! je
dvostruki faktorijel
)
∫
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}\,dx={\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
∞
e
−
a
x
sin
b
x
d
x
=
b
a
2
+
b
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\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\,dx={\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\,dx={\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\,dx={\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}
je modifikovana
Besselova funkcija
prve vrste)
∫
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)}
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