Nedoločeni integrali
uredi
Za integracijsko konstanto uporabljamo oznako C in jo lahko določimo, če je znana vrednost primitivne funkcije v neki točki. V splošnem pa je konstanta C nedoločena.
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
pri
n
≠
−
1
{\displaystyle \int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C\qquad {\mbox{ pri }}n\neq -1}
∫
x
−
1
d
x
=
∫
d
x
x
=
ln
|
x
|
+
C
{\displaystyle \int x^{-1}\,dx=\int {\frac {dx}{x}}=\ln {\left|x\right|}+C}
∫
x
d
x
=
2
3
x
3
/
2
+
C
{\displaystyle \int {\sqrt {x}}\,dx={\frac {2}{3}}x^{3/2}+C\!\,}
∫
1
x
d
x
=
2
x
+
C
{\displaystyle \int {\frac {1}{\sqrt {x}}}\,dx=2{\sqrt {x}}+C\!\,}
∫
1
1
−
x
2
d
x
=
arcsin
x
+
C
{\displaystyle \int {1 \over {\sqrt {1-x^{2}}}}\,dx=\arcsin {x}+C}
∫
1
a
−
x
2
d
x
=
arctan
x
a
−
x
2
+
C
{\displaystyle \int {1 \over {\sqrt {a-x^{2}}}}\,dx=\arctan {x \over {\sqrt {a-x^{2}}}}+C}
∫
x
x
2
−
a
d
x
=
x
2
−
a
+
C
{\displaystyle \int {x \over {\sqrt {x^{2}-a}}}\,dx={\sqrt {x^{2}-a}}\ +C}
∫
(
a
x
+
b
)
d
x
=
a
x
2
2
+
b
x
+
C
{\displaystyle \int (ax+b)\,dx={\frac {ax^{2}}{2}}+bx+C\!\,}
∫
(
a
x
2
+
b
x
+
c
)
d
x
=
a
3
x
3
+
b
2
x
2
+
c
x
+
C
{\displaystyle \int (ax^{2}+bx+c)\,dx={\frac {a}{3}}x^{3}+{\frac {b}{2}}x^{2}+cx+C\!\,}
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
{\displaystyle \int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\!\,}
∫
d
x
a
x
+
b
=
1
a
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {dx}{ax+b}}={\frac {1}{a}}\ln |ax+b|+C\!\,}
∫
1
x
2
+
1
d
x
=
arctan
x
+
C
{\displaystyle \int {\frac {1}{x^{2}+1}}\,dx=\arctan {x}+C}
∫
d
x
x
2
+
a
2
=
1
a
arctan
x
a
+
C
{\displaystyle \int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C\!\,}
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
|
f
(
x
)
|
+
C
{\displaystyle \int {\frac {f'(x)}{f(x)}}\,dx=\ln |f(x)|+C}
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}\,dx=e^{x}+C}
∫
e
c
x
d
x
=
1
c
e
c
x
+
C
{\displaystyle \int e^{cx}\;\mathrm {d} x={\frac {1}{c}}e^{cx}+C}
∫
a
x
d
x
=
a
x
ln
a
+
C
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+C}
∫
x
e
x
d
x
=
e
x
(
x
−
1
)
+
C
{\displaystyle \int xe^{x}\,dx=e^{x}(x-1)+C\!\,}
∫
d
x
e
x
=
−
1
e
x
+
C
{\displaystyle \int {\frac {dx}{e^{x}}}=-{\frac {1}{e^{x}}}+C\!\,}
∫
x
e
x
d
x
=
−
x
+
1
e
x
+
C
{\displaystyle \int {\frac {x}{e^{x}}}\,dx=-{\frac {x+1}{e^{x}}}+C\!\,}
∫
e
x
x
d
x
=
−
Ei
(
−
x
)
+
C
{\displaystyle \int {\frac {e^{x}}{x}}\,dx=-\operatorname {Ei} (-x)+C\!\,}
eksponentni integral
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
{\displaystyle \int \ln {x}\,dx=x\ln {x}-x+C}
∫
log
a
x
d
x
=
x
log
a
x
−
x
ln
a
+
C
{\displaystyle \int \log _{a}x\,dx=x\log _{a}{x}-{\frac {x}{\ln a}}+C}
∫
cos
(
n
x
)
d
x
=
s
i
n
(
n
x
)
n
+
C
{\displaystyle \int \cos({nx})\,dx={sin(nx) \over n}+C}
∫
sin
(
n
x
)
d
x
=
−
c
o
s
(
n
x
)
n
+
C
{\displaystyle \int \sin({nx})\,dx={-{cos(nx) \over n}}+C}
∫
tan
x
d
x
=
−
ln
|
cos
x
|
+
C
{\displaystyle \int \tan {x}\,dx=-\ln {\left|\cos {x}\right|}+C}
∫
cot
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int \cot {x}\,dx=\ln {\left|\sin {x}\right|}+C}
∫
d
x
cos
2
x
=
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int {\frac {dx}{\cos ^{2}x}}=\int \sec ^{2}x\,dx=\tan x+C}
∫
d
x
sin
2
x
=
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int {\frac {dx}{\sin ^{2}x}}=\int \csc ^{2}x\,dx=-\cot x+C}
∫
sin
2
x
d
x
=
2
x
−
sin
2
x
4
+
C
=
x
2
−
s
i
n
2
x
4
+
C
{\displaystyle \int \sin ^{2}x\,dx={2x-\sin 2x \over 4}+C={{x \over 2}-{sin2x \over 4}}+C}
∫
cos
2
x
d
x
=
2
x
+
sin
2
x
4
+
C
=
x
2
+
s
i
n
2
x
4
+
C
{\displaystyle \int \cos ^{2}x\,dx={2x+\sin 2x \over 4}+C={{x \over 2}+{sin2x \over 4}}+C}
∫
sinh
x
d
x
=
cosh
x
+
C
{\displaystyle \int \sinh x\,dx=\cosh x+C}
∫
cosh
x
d
x
=
sinh
x
+
C
{\displaystyle \int \cosh x\,dx=\sinh x+C}
∫
tanh
x
d
x
=
ln
|
cosh
x
|
+
C
{\displaystyle \int \tanh x\,dx=\ln |\cosh x|+C}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
C
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+C}
∫
csch
x
d
x
=
ln
|
tanh
x
2
|
+
C
{\displaystyle \int {\mbox{csch}}\,x\,dx=\ln \left|\tanh {x \over 2}\right|+C}
∫
sech
x
d
x
=
arctan
(
sinh
x
)
+
C
{\displaystyle \int {\mbox{sech}}\,x\,dx=\arctan(\sinh x)+C}
Določeni integrali
uredi
Obstajajo funkcije katerih primitivnih funkcij ne moremo izraziti v zaprti obliki. Vendar lahko izračunamo vrednosti določenih integralov teh funkcij v nekaterih intervalih . Nekaj uporabnih določenih integralov je podanih spodaj.
∫
0
∞
1
x
2
+
a
2
d
x
=
π
2
a
{\displaystyle \int _{0}^{\infty }{{\frac {1}{x^{2}+a^{2}}}\,dx}={\frac {\pi }{2a}}\!\,}
∫
0
∞
1
(
1
+
x
)
x
a
d
x
=
π
sin
(
a
π
)
,
(
a
<
1
)
{\displaystyle \int _{0}^{\infty }{{\frac {1}{(1+x)x^{a}}}\,dx}={\frac {\pi }{\sin(a\pi )}},\quad (a<1)\!\,}
∫
0
∞
x
a
−
1
1
+
x
d
x
=
π
sin
(
a
π
)
,
(
0
<
a
<
1
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{a-1}}{1+x}}\,dx}={\frac {\pi }{\sin(a\pi )}},\quad (0<a<1)\!\,}
∫
0
∞
1
(
1
−
x
)
x
a
d
x
=
−
π
ctg
,
(
a
<
1
)
{\displaystyle \int _{0}^{\infty }{{\frac {1}{(1-x)x^{a}}}\,dx}=-\pi \operatorname {ctg} ,\quad (a<1)\!\,}
∫
−
∞
∞
1
(
1
+
x
2
/
a
)
(
a
+
1
)
/
2
d
x
=
a
π
Γ
(
a
/
2
)
Γ
[
(
a
+
1
)
/
2
)
]
,
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{(1+x^{2}/a)^{(a+1)/2}}}\,dx={\frac {{\sqrt {a\pi }}\ \Gamma (a/2)}{\Gamma {\Big [}(a+1)/2){\Big ]}}},\quad (a>0)\!\,}
gostoto verjetnosti Studentove t-porazdelitve )
∫
0
∞
x
a
−
1
1
+
x
b
d
x
=
π
b
sin
(
a
π
/
b
)
,
(
0
<
a
<
b
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{a-1}}{1+x^{b}}}\,dx}={\frac {\pi }{b\sin(a\pi /b)}},\quad (0<a<b)\!\,}
∫
0
∞
x
a
x
b
+
c
b
d
x
=
π
c
a
+
1
−
b
b
sin
[
(
a
+
1
)
π
/
b
]
,
(
0
<
a
+
1
<
b
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{a}}{x^{b}+c^{b}}}\,dx}={\frac {\pi c^{a+1-b}}{b\sin {\Big [}(a+1)\pi /b{\Big ]}}},\quad (0<a+1<b)\!\,}
∫
0
∞
x
a
(
x
b
+
c
b
)
d
d
x
=
(
−
1
)
d
−
1
π
c
a
+
1
−
b
d
b
sin
[
(
a
+
1
)
π
/
b
]
(
d
−
1
)
!
Γ
[
(
a
+
1
)
/
(
b
−
d
+
1
)
]
,
(
0
<
a
+
1
<
b
d
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{a}}{(x^{b}+c^{b})^{d}}}\,dx}={\frac {(-1)^{d-1}\pi c^{a+1-bd}}{b\sin {\Big [}(a+1)\pi /b{\Big ]}(d-1)!\,\Gamma {\Big [}(a+1)/(b-d+1){\Big ]}}},\quad (0<a+1<bd)\!\,}
∫
0
∞
1
1
+
2
x
cos
(
a
)
+
x
2
d
x
=
a
sin
a
{\displaystyle \int _{0}^{\infty }{{\frac {1}{1+2x\,\cos(a)+x^{2}}}\,dx}={\frac {a}{\sin a}}\!\,}
∫
0
∞
x
b
1
+
2
x
cos
(
a
)
+
x
2
d
x
=
π
sin
(
b
π
)
sin
(
a
b
)
sin
a
,
(
0
<
a
<
π
2
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{b}}{1+2x\,\cos(a)+x^{2}}}\,dx}={\frac {\pi }{\sin(b\pi )}}{\frac {\sin(ab)}{\sin a}},\quad \left(0<a<{\frac {\pi }{2}}\right)\!\,}
∫
0
∞
1
e
a
x
d
x
=
1
a
,
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{ax}}}\,dx}={\frac {1}{a}},\quad (a>0)\!\,}
∫
−
∞
∞
1
e
a
x
2
d
x
=
2
∫
0
∞
1
e
a
x
2
d
x
=
π
a
,
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }{{\frac {1}{e^{ax^{2}}}}\,dx}=2\int _{0}^{\infty }{{\frac {1}{e^{ax^{2}}}}\,dx}={\sqrt {\frac {\pi }{a}}},\quad (a>0)\!\,}
Gaussov integral )
∫
0
∞
1
e
a
2
x
2
d
x
=
π
2
a
,
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{a^{2}x^{2}}}}\,dx}={\frac {\sqrt {\pi }}{2a}},\quad (a>0)\!\,}
∫
0
∞
1
e
x
2
+
(
a
2
/
x
2
)
d
x
=
π
2
e
2
a
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{x^{2}+(a^{2}/x^{2})}}}\,dx}={\frac {\sqrt {\pi }}{2e^{2a}}}\!\,}
∫
−
∞
∞
1
e
a
x
2
+
2
b
x
d
x
=
π
a
e
b
2
/
a
,
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }{{\frac {1}{e^{ax^{2}+2bx}}}\,dx}={\sqrt {\frac {\pi }{a}}}e^{b^{2}/a},\quad (a>0)\!\,}
∫
−
∞
∞
1
e
a
x
2
+
b
x
+
c
d
x
=
π
a
e
(
b
2
−
4
a
c
)
/
4
a
,
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }{{\frac {1}{e^{ax^{2}+bx+c}}}\,dx}={\frac {\sqrt {\pi }}{a}}e^{(b^{2}-4ac)/4a},\quad (a>0)\!\,}
∫
0
∞
x
e
x
2
d
x
=
1
2
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x^{2}}}}\,dx}={\frac {1}{2}}\!\,}
∫
0
∞
x
e
a
(
x
−
b
)
2
d
x
=
b
π
a
,
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{a(x-b)^{2}}}}\,dx}=b{\sqrt {\frac {\pi }{a}}},\quad (a>0)\!\,}
∫
0
∞
x
2
e
a
x
2
d
x
=
1
4
π
a
3
,
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{2}}{e^{ax^{2}}}}\,dx}={\frac {1}{4}}{\sqrt {\frac {\pi }{a^{3}}}},\quad (a>0)\!\,}
∫
0
∞
x
2
e
a
2
x
2
d
x
=
π
4
a
3
,
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{2}}{e^{a^{2}x^{2}}}}\,dx}={\frac {\sqrt {\pi }}{4a^{3}}},\quad (a>0)\!\,}
∫
0
∞
x
n
e
a
x
d
x
=
{
Γ
(
n
+
1
)
a
n
+
1
,
(
a
>
0
,
n
>
−
1
)
n
!
a
n
+
1
,
(
a
>
0
,
n
≥
0
;
(
n
∈
N
0
)
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{n}}{e^{ax}}}\,dx}={\begin{cases}{\frac {\Gamma (n+1)}{a^{n+1}}},&\ (a>0,n>-1)\\{\frac {n!}{a^{n+1}}},&\ (a>0,n\geq 0;\,(n\in \mathbb {N} _{0}))\end{cases}}\!\,}
∫
0
∞
x
n
e
a
x
2
d
x
=
{
Γ
[
(
n
+
1
)
/
2
]
2
a
(
n
+
1
)
/
2
,
(
a
>
0
,
n
>
−
1
)
(
2
k
−
1
)
!
!
2
k
+
1
a
k
,
(
a
>
0
,
n
=
2
k
,
k
∈
Z
)
k
!
2
a
k
+
1
,
(
a
>
0
,
n
=
2
k
+
1
,
k
∈
Z
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{n}}{e^{ax^{2}}}}\,dx}={\begin{cases}{\frac {\Gamma {\big [}(n+1)/2{\big ]}}{2a^{(n+1)/2}}},&\ (a>0,n>-1)\\{\frac {(2k-1)!!}{2^{k+1}a^{k}}},&\ (a>0,n=2k,k\in \mathbb {Z} )\\{\frac {k!}{2a^{k+1}}},&\ (a>0,n=2k+1,k\in \mathbb {Z} )\end{cases}}\!\,}
dvojna fakulteta )
∫
0
∞
x
2
n
e
a
x
2
d
x
=
1
⋅
3
⋅
5
⋯
(
2
n
−
1
)
2
n
+
1
a
n
π
a
{\displaystyle \int _{0}^{\infty }{{\frac {x^{2n}}{e^{ax^{2}}}}\,dx}={\frac {1\cdot 3\cdot 5\cdots (2n-1)}{2^{n+1}a^{n}}}{\sqrt {\frac {\pi }{a}}}\!\,}
∫
0
∞
x
2
n
+
1
e
a
x
2
d
x
=
n
!
2
a
n
+
1
,
(
a
>
0
,
n
>
−
1
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{2n+1}}{e^{ax^{2}}}}\,dx}={\frac {n!}{2a^{n+1}}},\quad (a>0,n>-1)\!\,}
∫
0
∞
x
n
n
e
x
d
x
=
Γ
(
n
)
{\displaystyle \int _{0}^{\infty }{\frac {x^{n}}{ne^{x}}}\,dx=\Gamma (n)\!\,}
funkcija Γ )
∫
0
∞
ln
x
e
x
d
x
=
−
γ
=
−
0
,
5772156649
…
{\displaystyle \int _{0}^{\infty }{\frac {\ln x}{e^{x}}}\,dx=-\gamma =-0,5772156649\ldots \!\,}
Euler-Mascheronijeva konstanta )
∫
0
∞
x
e
a
x
d
x
=
1
2
a
π
a
{\displaystyle \int _{0}^{\infty }{{\frac {\sqrt {x}}{e^{ax}}}\,dx}={\frac {1}{2a}}{\sqrt {\frac {\pi }{a}}}\!\,}
∫
0
∞
1
e
x
x
d
x
=
Γ
(
1
2
)
=
π
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{x}{\sqrt {x}}}}\,dx}=\Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}\!\,}
Eulerjev integral )
∫
0
∞
1
e
a
x
x
d
x
=
π
a
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{ax}{\sqrt {x}}}}\,dx}={\sqrt {\frac {\pi }{a}}}\!\,}
∫
0
∞
x
e
x
−
1
d
x
=
Γ
(
2
)
ζ
(
2
)
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}=\Gamma (2)\zeta (2)={\frac {\pi ^{2}}{6}}}
ζ
(
⋅
)
{\displaystyle \zeta (\cdot )}
Riemannova funkcija ζ (baselski problem ))
∫
0
∞
x
3
e
x
−
1
d
x
=
Γ
(
4
)
ζ
(
4
)
=
π
4
15
{\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{x}-1}}\,dx}=\Gamma (4)\zeta (4)={\frac {\pi ^{4}}{15}}}
∫
0
∞
x
n
e
x
−
1
d
x
=
Γ
(
n
+
1
)
ζ
(
n
+
1
)
{\displaystyle \int _{0}^{\infty }{{\frac {x^{n}}{e^{x}-1}}\,dx}=\Gamma (n+1)\zeta (n+1)}
∫
0
∞
x
ln
x
e
2
π
x
−
1
d
x
=
1
24
−
1
2
ln
A
{\displaystyle \int _{0}^{\infty }{\frac {x\ln x}{e^{2\pi x}-1}}\,dx={\frac {1}{24}}-{\frac {1}{2}}\ln A\!\,}
A je Glaisher-Kinkelinova konstanta )
∫
0
1
/
2
ln
Γ
(
x
+
1
)
d
x
=
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1
2
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7
24
ln
2
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1
4
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π
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3
2
ln
A
{\displaystyle \int _{0}^{1/2}\ln \Gamma (x+1)\,dx=-{\frac {1}{2}}-{\frac {7}{24}}\ln 2+{\frac {1}{4}}\ln \pi +{\frac {3}{2}}\ln A\!\,}
![{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{(1+x^{2}/a)^{(a+1)/2}}}\,dx={\frac {{\sqrt {a\pi }}\ \Gamma (a/2)}{\Gamma {\Big [}(a+1)/2){\Big ]}}},\quad (a>0)\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/595d15fc8a486c9945439b1f4d378956b3df5838)
![{\displaystyle \int _{0}^{\infty }{{\frac {x^{a}}{x^{b}+c^{b}}}\,dx}={\frac {\pi c^{a+1-b}}{b\sin {\Big [}(a+1)\pi /b{\Big ]}}},\quad (0<a+1<b)\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02fcd693dc6b4637226b053f4ec38c7cbe9ffffb)
![{\displaystyle \int _{0}^{\infty }{{\frac {x^{a}}{(x^{b}+c^{b})^{d}}}\,dx}={\frac {(-1)^{d-1}\pi c^{a+1-bd}}{b\sin {\Big [}(a+1)\pi /b{\Big ]}(d-1)!\,\Gamma {\Big [}(a+1)/(b-d+1){\Big ]}}},\quad (0<a+1<bd)\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2b7726d996ed426bce5a35e81db9abbd3ac504d)
![{\displaystyle \int _{0}^{\infty }{{\frac {x^{n}}{e^{ax^{2}}}}\,dx}={\begin{cases}{\frac {\Gamma {\big [}(n+1)/2{\big ]}}{2a^{(n+1)/2}}},&\ (a>0,n>-1)\\{\frac {(2k-1)!!}{2^{k+1}a^{k}}},&\ (a>0,n=2k,k\in \mathbb {Z} )\\{\frac {k!}{2a^{k+1}}},&\ (a>0,n=2k+1,k\in \mathbb {Z} )\end{cases}}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95e7326d03d2cd5229a04e9fa92cf7db34327635)
