Інтегрування є одною з двох основних операцій математичного аналізу . Тоді як диференціювання має прості правила, за якими можна знайти похідну складних функцій через диференціювання її складових функцій, для інтегралів це не так, і тому таблиці відомих первісних виявляються часто дуже корисними. На цій сторінці представлено список основних первісних.
C вживається як довільна стала інтегрування інтегрування, яку можна визначити якщо відомо значення інтеграла в якій-небудь точці.
ред.
∫
c
f
(
x
)
d
x
=
c
∫
f
(
x
)
d
x
{\displaystyle \int cf(x)\,dx=c\int f(x)\,dx}
∫
f
(
a
x
+
b
)
d
x
=
1
a
F
(
a
x
+
b
)
+
C
{\displaystyle \int f(ax+b)\,dx={1 \over a}F(ax+b)\,+C}
∫
[
f
(
x
)
+
g
(
x
)
]
d
x
=
∫
f
(
x
)
d
x
+
∫
g
(
x
)
d
x
{\displaystyle \int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}
∫
[
f
(
x
)
−
g
(
x
)
]
d
x
=
∫
f
(
x
)
d
x
−
∫
g
(
x
)
d
x
{\displaystyle \int [f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx}
∫
f
(
x
)
g
(
x
)
d
x
=
f
(
x
)
∫
g
(
x
)
d
x
−
∫
(
d
[
f
(
x
)
]
∫
g
(
x
)
d
x
)
d
x
{\displaystyle \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left(d[f(x)]\int g(x)\,dx\right)\,dx}
∫
u
(
x
)
v
′
(
x
)
d
x
=
u
(
x
)
v
(
x
)
−
∫
u
′
(
x
)
v
(
x
)
d
x
{\displaystyle \int u(x)v'(x)\,dx=u(x)v(x)-\int u'(x)v(x)\,dx}
ред.
∫
d
x
=
x
+
C
{\displaystyle \int \,dx=x+C}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
,
{\displaystyle \int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C,}
n
≠
−
1
{\displaystyle n\neq -1}
∫
d
x
x
=
ln
|
x
|
+
C
{\displaystyle \int {\frac {dx}{x}}\,=\ln {\left|x\right|}+C}
∫
d
x
a
2
+
x
2
=
1
a
arctg
x
a
+
C
{\displaystyle \int {dx \over {a^{2}+x^{2}}}={1 \over a}\operatorname {arctg} {x \over a}+C}
∫
d
x
x
2
−
a
2
=
1
2
a
ln
|
x
−
a
x
+
a
|
+
C
{\displaystyle \int \!{dx \over {x^{2}-a^{2}}}={1 \over 2a}\ln \left|{x-a \over {x+a}}\right|+C}
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
{\displaystyle \int \ln {x}\,dx=x\ln {x}-x+C}
∫
log
b
x
d
x
=
x
log
b
x
−
x
log
b
e
+
C
{\displaystyle \int \log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+C}
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}\,dx=e^{x}+C}
∫
a
x
d
x
=
a
x
ln
a
+
C
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+C}
∫
d
x
a
2
−
x
2
=
arcsin
x
a
+
C
{\displaystyle \int {dx \over {\sqrt {a^{2}-x^{2}}}}=\arcsin {x \over a}+C}
∫
−
d
x
a
2
−
x
2
=
arccos
x
a
+
C
{\displaystyle \int {-dx \over {\sqrt {a^{2}-x^{2}}}}=\arccos {x \over a}+C}
∫
d
x
x
x
2
−
a
2
=
1
a
arcsec
|
x
|
a
+
C
{\displaystyle \int {dx \over x{\sqrt {x^{2}-a^{2}}}}={1 \over a}{\mbox{arcsec}}\,{|x| \over a}+C}
∫
d
x
x
2
+
a
2
=
ln
|
x
+
x
2
+
a
2
|
+
C
{\displaystyle \int \!{dx \over {\sqrt {x^{2}+a^{2}}}}=\ln \left|{x+{\sqrt {x^{2}+a^{2}}}}\right|+C}
∫
d
x
x
2
−
a
2
=
ln
|
x
+
x
2
−
a
2
|
+
C
{\displaystyle \int \!{dx \over {\sqrt {x^{2}-a^{2}}}}=\ln \left|{x+{\sqrt {x^{2}-a^{2}}}}\right|+C}
∫
sin
x
d
x
=
−
cos
x
+
C
{\displaystyle \int \sin {x}\,dx=-\cos {x}+C}
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos {x}\,dx=\sin {x}+C}
∫
tg
x
d
x
=
−
ln
|
cos
x
|
+
C
{\displaystyle \int \operatorname {tg} x\,dx=-\ln {\left|\cos {x}\right|}+C}
∫
ctg
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int \operatorname {ctg} x\,dx=\ln {\left|\sin {x}\right|}+C}
∫
sec
x
d
x
=
ln
|
sec
x
+
tg
x
|
+
C
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\operatorname {tg} x\right|}+C}
∫
cosec
x
d
x
=
−
ln
|
cosec
x
+
ctg
x
|
+
C
{\displaystyle \int \operatorname {cosec} x\,dx=-\ln {\left|\operatorname {cosec} x+\operatorname {ctg} x\right|}+C}
∫
sec
2
x
d
x
=
tg
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\operatorname {tg} x+C}
∫
cosec
2
x
d
x
=
−
ctg
x
+
C
{\displaystyle \int \operatorname {cosec} ^{2}x\,dx=-\operatorname {ctg} x+C}
∫
sec
x
tg
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec {x}\,\operatorname {tg} x\,dx=\sec {x}+C}
∫
cosec
x
ctg
x
d
x
=
−
cosec
x
+
C
{\displaystyle \int \operatorname {cosec} x\,\operatorname {ctg} x\,dx=-\operatorname {cosec} x+C}
∫
sin
2
x
d
x
=
1
2
(
x
−
sin
x
cos
x
)
+
C
{\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}(x-\sin x\cos x)+C}
∫
cos
2
x
d
x
=
1
2
(
x
+
sin
x
cos
x
)
+
C
{\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}(x+\sin x\cos x)+C}
∫
sec
3
x
d
x
=
1
2
sec
x
tg
x
+
1
2
ln
|
sec
x
+
tg
x
|
+
C
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}\sec x\operatorname {tg} x+{\frac {1}{2}}\ln |\sec x+\operatorname {tg} x|+C}
∫
sin
n
x
d
x
=
−
sin
n
−
1
x
cos
x
n
+
n
−
1
n
∫
sin
n
−
2
x
d
x
{\displaystyle \int \sin ^{n}x\,dx=-{\frac {\sin ^{n-1}{x}\cos {x}}{n}}+{\frac {n-1}{n}}\int \sin ^{n-2}{x}\,dx}
∫
cos
n
x
d
x
=
−
cos
n
−
1
x
sin
x
n
+
n
−
1
n
∫
cos
n
−
2
x
d
x
{\displaystyle \int \cos ^{n}x\,dx=-{\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \cos ^{n-2}{x}\,dx}
∫
d
x
cos
2
x
=
tg
x
+
C
{\displaystyle \int \!{dx \over \cos ^{2}x}=\operatorname {tg} \,x+C}
∫
d
x
sin
2
x
=
−
ctg
x
+
C
{\displaystyle \int \!{dx \over \sin ^{2}x}=-\operatorname {ctg} \,x+C}
ред.
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
C
{\displaystyle \int \arcsin {x}\,dx=x\,\arcsin {x}+{\sqrt {1-x^{2}}}+C}
∫
arccos
x
d
x
=
x
arccos
x
−
1
−
x
2
+
C
{\displaystyle \int \arccos {x}\,dx=x\,\arccos {x}-{\sqrt {1-x^{2}}}+C}
∫
arctg
x
d
x
=
x
arctg
x
−
1
2
ln
|
1
+
x
2
|
+
C
{\displaystyle \int \operatorname {arctg} x\,dx=x\,\operatorname {arctg} x-{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+C}
∫
arcsec
x
d
x
=
x
arcsec
x
+
x
2
−
1
ln
(
x
+
x
2
−
1
)
x
1
−
1
x
2
+
C
{\displaystyle \int \operatorname {arcsec} {x}\,dx=x\,\operatorname {arcsec} {x}+{\frac {{\sqrt {x^{2}-1}}\ln {(x+{\sqrt {x^{2}-1}})}}{x\,{\sqrt {1-{\frac {1}{x^{2}}}}}}}+C}
∫
arccosec
x
d
x
=
x
arccosec
x
+
x
2
−
1
ln
(
x
+
x
2
−
1
)
x
1
−
1
x
2
+
C
{\displaystyle \int \operatorname {arccosec} x\,dx=x\,\operatorname {arccosec} x+{\frac {{\sqrt {x^{2}-1}}\ln {(x+{\sqrt {x^{2}-1}})}}{x\,{\sqrt {1-{\frac {1}{x^{2}}}}}}}+C}
∫
arcctg
x
d
x
=
x
arcctg
x
+
1
2
ln
|
1
+
x
2
|
+
C
{\displaystyle \int \operatorname {arcctg} x\,dx=x\,\operatorname {arcctg} x+{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+C}
∫
sh
x
d
x
=
ch
x
+
C
{\displaystyle \int \operatorname {sh} x\,dx=\operatorname {ch} x+C}
∫
ch
x
d
x
=
sh
x
+
C
{\displaystyle \int \operatorname {ch} x\,dx=\operatorname {sh} x+C}
∫
th
x
d
x
=
ln
|
ch
x
|
+
C
{\displaystyle \int \operatorname {th} x\,dx=\ln |\operatorname {ch} x|+C}
∫
csch
x
d
x
=
ln
|
th
x
2
|
+
C
{\displaystyle \int \operatorname {csch} x\,dx=\ln \left|\operatorname {th} {x \over 2}\right|+C}
∫
sech
x
d
x
=
arctg
(
sh
x
)
+
C
{\displaystyle \int \operatorname {sech} x\,dx=\operatorname {arctg} (\operatorname {sh} x)+C}
∫
cth
x
d
x
=
ln
|
sh
x
|
+
C
{\displaystyle \int \operatorname {cth} x\,dx=\ln |\operatorname {sh} x|+C}
∫
sech
2
x
d
x
=
th
x
+
C
{\displaystyle \int \operatorname {sech} ^{2}x\,dx=\operatorname {th} x+C}
ред.
∫
arsh
x
d
x
=
x
arsh
x
−
x
2
+
1
+
C
{\displaystyle \int \operatorname {arsh} \,x\,dx=x\,\operatorname {arsh} \,x-{\sqrt {x^{2}+1}}+C}
∫
arch
x
d
x
=
x
arch
x
−
x
2
−
1
+
C
{\displaystyle \int \operatorname {arch} \,x\,dx=x\,\operatorname {arch} \,x-{\sqrt {x^{2}-1}}+C}
∫
arth
x
d
x
=
x
arth
x
+
1
2
ln
(
1
−
x
2
)
+
C
{\displaystyle \int \operatorname {arth} \,x\,dx=x\,\operatorname {arth} \,x+{\frac {1}{2}}\ln {(1-x^{2})}+C}
∫
arcsch
x
d
x
=
x
arcsch
x
+
ln
[
x
(
1
+
1
x
2
+
1
)
]
+
C
{\displaystyle \int \operatorname {arcsch} \,x\,dx=x\,\operatorname {arcsch} \,x+\ln {\left[x\left({\sqrt {1+{\frac {1}{x^{2}}}}}+1\right)\right]}+C}
∫
arsech
x
d
x
=
x
arsech
x
−
arctg
(
x
x
−
1
1
−
x
1
+
x
)
+
C
{\displaystyle \int \operatorname {arsech} \,x\,dx=x\,\operatorname {arsech} \,x-\operatorname {arctg} {\left({\frac {x}{x-1}}{\sqrt {\frac {1-x}{1+x}}}\right)}+C}
∫
arcth
x
d
x
=
x
arcth
x
+
1
2
ln
(
x
2
−
1
)
+
C
{\displaystyle \int \operatorname {arcth} \,x\,dx=x\,\operatorname {arcth} \,x+{\frac {1}{2}}\ln {(x^{2}-1)}+C}
∫
cos
a
x
e
b
x
d
x
=
e
b
x
a
2
+
b
2
(
a
sin
a
x
+
b
cos
a
x
)
+
C
{\displaystyle \int \cos ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(a\sin ax+b\cos ax\right)+C}
∫
sin
a
x
e
b
x
d
x
=
e
b
x
a
2
+
b
2
(
b
sin
a
x
−
a
cos
a
x
)
+
C
{\displaystyle \int \sin ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(b\sin ax-a\cos ax\right)+C}
∫
cos
a
x
ch
b
x
d
x
=
1
a
2
+
b
2
(
a
sin
a
x
ch
b
x
+
b
cos
a
x
sh
b
x
)
+
C
{\displaystyle \int \cos ax\,\operatorname {ch} bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(a\sin ax\,\operatorname {ch} bx+b\cos ax\,\operatorname {sh} bx\right)+C}
∫
sin
a
x
ch
b
x
d
x
=
1
a
2
+
b
2
(
b
sin
a
x
sh
b
x
−
a
cos
a
x
ch
b
x
)
+
C
{\displaystyle \int \sin ax\,\operatorname {ch} bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(b\sin ax\,\operatorname {sh} bx-a\cos ax\,\operatorname {ch} bx\right)+C}
ред.
∫
|
(
a
x
+
b
)
n
|
d
x
=
(
a
x
+
b
)
n
+
2
a
(
n
+
1
)
|
a
x
+
b
|
+
C
[
n
is odd, and
n
≠
−
1
]
{\displaystyle \int \left|(ax+b)^{n}\right|\,dx={(ax+b)^{n+2} \over a(n+1)\left|ax+b\right|}+C\,\,[\,n{\text{ is odd, and }}n\neq -1\,]}
∫
|
sin
a
x
|
d
x
=
−
1
a
|
sin
a
x
|
ctg
a
x
+
C
{\displaystyle \int \left|\sin {ax}\right|\,dx={-1 \over a}\left|\sin {ax}\right|\operatorname {ctg} {ax}+C}
∫
|
cos
a
x
|
d
x
=
1
a
|
cos
a
x
|
tg
a
x
+
C
{\displaystyle \int \left|\cos {ax}\right|\,dx={1 \over a}\left|\cos {ax}\right|\operatorname {tg} {ax}+C}
∫
|
tg
a
x
|
d
x
=
tg
a
x
[
−
ln
|
cos
a
x
|
]
a
|
tg
a
x
|
+
C
{\displaystyle \int \left|\operatorname {tg} {ax}\right|\,dx={\operatorname {tg} {ax}[-\ln \left|\cos {ax}\right|] \over a\left|\operatorname {tg} {ax}\right|}+C}
∫
|
cosec
a
x
|
d
x
=
−
ln
|
cosec
a
x
+
ctg
a
x
|
sin
a
x
a
|
sin
a
x
|
+
C
{\displaystyle \int \left|\operatorname {cosec} {ax}\right|\,dx={-\ln \left|\operatorname {cosec} {ax}+\operatorname {ctg} {ax}\right|\sin {ax} \over a\left|\sin {ax}\right|}+C}
∫
|
sec
a
x
|
d
x
=
ln
|
sec
a
x
+
tg
a
x
|
cos
a
x
a
|
cos
a
x
|
+
C
{\displaystyle \int \left|\sec {ax}\right|\,dx={\ln \left|\sec {ax}+\operatorname {tg} {ax}\right|\cos {ax} \over a\left|\cos {ax}\right|}+C}
∫
|
ctg
a
x
|
d
x
=
tg
a
x
[
ln
|
sin
a
x
|
]
a
|
tg
a
x
|
+
C
{\displaystyle \int \left|\operatorname {ctg} {ax}\right|\,dx={\operatorname {tg} {ax}[\ln \left|\sin {ax}\right|] \over a\left|\operatorname {tg} {ax}\right|}+C}
∫
Ci
(
x
)
d
x
=
x
Ci
(
x
)
−
sin
x
{\displaystyle \int \operatorname {Ci} (x)dx=x\,\operatorname {Ci} (x)-\sin x}
∫
Si
(
x
)
d
x
=
x
Si
(
x
)
+
cos
x
{\displaystyle \int \operatorname {Si} (x)dx=x\,\operatorname {Si} (x)+\cos x}
∫
Ei
(
x
)
d
x
=
x
Ei
(
x
)
−
e
x
{\displaystyle \int \operatorname {Ei} (x)dx=x\,\operatorname {Ei} (x)-e^{x}}
∫
li
(
x
)
d
x
=
x
li
(
x
)
−
Ei
(
2
ln
x
)
{\displaystyle \int \operatorname {li} (x)dx=x\,\operatorname {li} (x)-\operatorname {Ei} (2\ln x)}
∫
li
(
x
)
x
d
x
=
ln
x
li
(
x
)
−
x
{\displaystyle \int {\frac {\operatorname {li} (x)}{x}}\,dx=\ln x\,\operatorname {li} (x)-x}
∫
erf
(
x
)
d
x
=
e
−
x
2
π
+
x
erf
(
x
)
{\displaystyle \int \operatorname {erf} (x)\,dx={\frac {e^{-x^{2}}}{\sqrt {\pi }}}+x\,{\text{erf}}(x)}
ред.
Для деяких функцій, чиї первісні не можуть бути представлені явно, тим не менш їхні деякі визначені інтеграли можуть бути обчислені. Тут перелічені деякі популярні інтеграли
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
Гамма-функція )
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
Гаусовий інтеграл )
∫
0
∞
x
e
−
a
2
x
2
d
x
=
1
2
a
2
{\displaystyle \int _{0}^{\infty }{xe^{-a^{2}x^{2}}\,dx}={\frac {1}{2a^{2}}}}
∫
0
∞
x
2
e
−
a
2
x
2
d
x
=
π
4
a
3
{\displaystyle \int _{0}^{\infty }{x^{2}e^{-a^{2}x^{2}}\,dx}={\frac {\sqrt {\pi }}{4a^{3}}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
x
2
i
+
1
e
−
a
2
x
2
d
x
=
a
!
2
a
2
i
+
2
{\displaystyle \int _{0}^{\infty }{x^{2i+1}e^{-a^{2}x^{2}}\,dx}={\frac {a!}{2a^{2i+2}}}}
a
>
0
;
i
=
1
,
2
,
3...
{\displaystyle a>0;\;\;\!i=1,2,3...}
∫
0
∞
x
2
i
e
−
a
2
x
2
d
x
=
1
∗
3
∗
5
∗
.
.
.
∗
(
2
i
−
1
)
2
i
+
1
a
2
i
+
1
π
{\displaystyle \int _{0}^{\infty }{x^{2i}e^{-a^{2}x^{2}}\,dx}={\frac {1*3*5*...*(2i-1)}{2^{i+1}a^{2i+1}}}{\sqrt {\pi }}}
a
>
0
;
i
=
1
,
2
,
3...
{\displaystyle a>0;\;\;\!i=1,2,3...}
∫
0
∞
x
n
e
−
a
2
x
2
d
x
=
Γ
(
n
+
1
2
)
2
a
n
+
1
{\displaystyle \int _{0}^{\infty }{x^{n}e^{-a^{2}x^{2}}\,dx}={\frac {\Gamma ({\frac {n+1}{2}})}{2a^{n+1}}}}
a
>
0
{\displaystyle a>0\!}
Гамма-функція )
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
числа Бернуллі )
∫
0
∞
x
3
e
x
−
1
d
x
=
π
4
15
{\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{x}-1}}\,dx}={\frac {\pi ^{4}}{15}}}
∫
0
∞
1
e
a
x
+
1
d
x
=
ln
2
a
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{ax}+1}}\,dx}={\frac {\ln {2}}{a}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
x
e
a
x
+
1
d
x
=
π
2
12
a
2
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{ax}+1}}\,dx}={\frac {\pi ^{2}}{12a^{2}}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
x
3
e
a
x
+
1
d
x
=
7
120
π
4
a
4
{\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{ax}+1}}\,dx}={\frac {7}{120}}{\frac {\pi ^{4}}{a^{4}}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
1
⋅
3
⋅
5
⋅
⋯
⋅
(
n
−
1
)
2
⋅
4
⋅
6
⋅
⋯
⋅
n
π
2
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}{x}\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}{x}\,dx={\frac {1\cdot 3\cdot 5\cdot \cdots \cdot (n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot n}}{\frac {\pi }{2}}}
n парне число і
n
≥
2
{\displaystyle \scriptstyle {n\geq 2}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
2
⋅
4
⋅
6
⋅
⋯
⋅
(
n
−
1
)
3
⋅
5
⋅
7
⋅
⋯
⋅
n
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}{x}\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}{x}\,dx={\frac {2\cdot 4\cdot 6\cdot \cdots \cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdots \cdot n}}}
n
{\displaystyle \scriptstyle {n}}
n
≥
3
{\displaystyle \scriptstyle {n\geq 3}}
∫
−
π
π
cos
(
α
x
)
cos
n
(
β
x
)
d
x
=
{
2
π
2
n
(
n
m
)
|
α
|
=
|
β
(
2
m
−
n
)
|
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\cos ^{n}(\beta x)dx=\left\{{\begin{array}{cc}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&|\alpha |=|\beta (2m-n)|\\0&{\mbox{otherwise}}\\\end{array}}\right.}
α
,
β
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β
≠
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥
0
{\displaystyle \scriptstyle m,n\geq 0}
Біноміальний коефіцієнт )
∫
−
π
π
sin
(
α
x
)
cos
n
(
β
x
)
d
x
=
0
{\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\cos ^{n}(\beta x)dx=0}
α
,
β
{\displaystyle \scriptstyle \alpha ,\beta }
n
{\displaystyle \scriptstyle n}
Симетрія )
∫
−
π
π
sin
(
α
x
)
sin
n
(
β
x
)
d
x
=
{
(
−
1
)
(
n
+
1
)
/
2
(
−
1
)
m
2
π
2
n
(
n
m
)
n
odd
,
α
=
β
(
2
m
−
n
)
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\sin ^{n}(\beta x)dx=\left\{{\begin{array}{cc}(-1)^{(n+1)/2}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\mbox{ odd}},\ \alpha =\beta (2m-n)\\0&{\mbox{otherwise}}\\\end{array}}\right.}
α
,
β
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β
≠
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥
0
{\displaystyle \scriptstyle m,n\geq 0}
Біноміальний коефіцієнт )
∫
−
π
π
cos
(
α
x
)
sin
n
(
β
x
)
d
x
=
{
(
−
1
)
n
/
2
(
−
1
)
m
2
π
2
n
(
n
m
)
n
even
,
|
α
|
=
|
β
(
2
m
−
n
)
|
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\sin ^{n}(\beta x)dx=\left\{{\begin{array}{cc}(-1)^{n/2}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\mbox{ even}},\ |\alpha |=|\beta (2m-n)|\\0&{\mbox{otherwise}}\\\end{array}}\right.}
α
,
β
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β
≠
0
{\displaystyle \scriptstyle \beta \neq 0\!}
m
,
n
≥
0
{\displaystyle \scriptstyle m,n\geq 0\!}
Біноміальний коефіцієнт )
∫
0
∞
sin
2
x
x
2
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}}
∫
0
∞
x
z
−
1
e
−
x
d
x
=
Γ
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
Γ
(
z
)
{\displaystyle \Gamma (z)\!}
Гамма-функція )
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π
a
exp
[
b
2
−
4
a
c
4
a
]
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}
exp
[
u
]
{\displaystyle \exp[u]\!}
експонента
e
u
{\displaystyle e^{u}}
a
>
0
{\displaystyle 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
(
x
)
{\displaystyle I_{0}(x)\!}
Функція Бесселя першого роду)
∫
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)}
∫
−
∞
∞
(
1
+
x
2
/
ν
)
−
(
ν
+
1
)
/
2
d
x
=
ν
π
Γ
(
ν
/
2
)
Γ
(
(
ν
+
1
)
/
2
)
)
{\displaystyle \int _{-\infty }^{\infty }{(1+x^{2}/\nu )^{-(\nu +1)/2}dx}={\frac {{\sqrt {\nu \pi }}\ \Gamma (\nu /2)}{\Gamma ((\nu +1)/2))}}\,}
ν
>
0
{\displaystyle \nu >0\,}
густини ймовірності для T-розподілу Стьюдента Для загального випадку, якщо первісної не існує, застосовується метод вичерпання :
∫
a
b
f
(
x
)
d
x
=
(
b
−
a
)
∑
n
=
1
∞
∑
m
=
1
2
n
−
1
(
−
1
)
m
+
1
2
−
n
f
(
a
+
m
(
b
−
a
)
2
−
n
)
.
{\displaystyle \int _{a}^{b}{f(x)\,dx}=(b-a)\sum \limits _{n=1}^{\infty }{\sum \limits _{m=1}^{2^{n}-1}{\left({-1}\right)^{m+1}}}2^{-n}f(a+m\left({b-a}\right)2^{-n}).}
∫
0
1
[
ln
(
1
/
x
)
]
p
d
x
=
p
!
{\displaystyle \int _{0}^{1}[\ln(1/x)]^{p}\,dx=p!}
ред.
∫
0
1
x
−
x
d
x
=
∑
n
=
1
∞
n
−
n
(
=
1.29128599706266
…
)
∫
0
1
x
x
d
x
=
−
∑
n
=
1
∞
(
−
1
)
n
n
−
n
(
=
0.783430510712
…
)
{\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}&&(=1.29128599706266\dots )\\\int _{0}^{1}x^{x}\,dx&=-\sum _{n=1}^{\infty }(-1)^{n}n^{-n}&&(=0.783430510712\dots )\end{aligned}}}
Обчислені Йоганном Бернуллі .
![{\displaystyle \int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dc2c6e2acadf432d22ca42bc6a21af25e48e64d)
![{\displaystyle \int [f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5c5f0ce4c12f5715703d07f46dbd9e067acdd3f)
![{\displaystyle \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left(d[f(x)]\int g(x)\,dx\right)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2da09fdfdcb12ff1a55cb20340291b01e6832c0)
![{\displaystyle \int \operatorname {arcsch} \,x\,dx=x\,\operatorname {arcsch} \,x+\ln {\left[x\left({\sqrt {1+{\frac {1}{x^{2}}}}}+1\right)\right]}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ca1160efc8b0a76d64c89bca2625a9c7828dd9f)
![{\displaystyle \int \left|(ax+b)^{n}\right|\,dx={(ax+b)^{n+2} \over a(n+1)\left|ax+b\right|}+C\,\,[\,n{\text{ is odd, and }}n\neq -1\,]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9db551fb843fee76fb6a9325999ab8c0515a8d1f)
![{\displaystyle \int \left|\operatorname {tg} {ax}\right|\,dx={\operatorname {tg} {ax}[-\ln \left|\cos {ax}\right|] \over a\left|\operatorname {tg} {ax}\right|}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1e4818bb015836e63dae22391b4281d417d0f84)
![{\displaystyle \int \left|\operatorname {ctg} {ax}\right|\,dx={\operatorname {tg} {ax}[\ln \left|\sin {ax}\right|] \over a\left|\operatorname {tg} {ax}\right|}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21bc729bf3f639ad0e77341177afd98344e4140d)
![{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9aac3de5d20ec70a35c8d52bafaa6a82c148b566)
![{\displaystyle \exp[u]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/432e3d08f86cd0698a9594d565526b7ebdef5c5a)
![{\displaystyle \int _{0}^{1}[\ln(1/x)]^{p}\,dx=p!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e5028566f81ea22ae449458f859edfc08a532cd)
