Список интегралов элементарных функций
Интегрирование — это одна из двух основных операций в математическом анализе . В отличие от операции дифференцирования, интеграл от элементарной функции может не быть элементарной функцией. Например, из теоремы Лиувилля следует, что интеграл от
e
x
2
{\displaystyle e^{x^{2}}}
первообразных оказываются часто очень полезны, хотя сейчас и теряют свою актуальность с появлением систем компьютерной алгебры.
На этой странице представлен список наиболее часто встречающихся первообразных.
C
{\displaystyle C}
∫ ∫ -->
C
f
(
x
)
d
x
=
C
∫ ∫ -->
f
(
x
)
d
x
{\displaystyle \int Cf(x)\,dx=C\int f(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
− − -->
∫ ∫ -->
(
∫ ∫ -->
g
(
x
)
d
x
)
d
f
(
x
)
{\displaystyle \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left(\int g(x)\,dx\right)\,df(x)}
∫ ∫ -->
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}
∫ ∫ -->
0
d
x
=
C
{\displaystyle \int \!0\,dx=C}
∫ ∫ -->
a
d
x
=
a
x
+
C
{\displaystyle \int \!a\,dx=ax+C}
∫ ∫ -->
x
n
d
x
=
{
x
n
+
1
n
+
1
+
C
,
n
≠ ≠ -->
− − -->
1
ln
-->
|
x
|
+
C
,
n
=
− − -->
1
{\displaystyle \int \!x^{n}\,dx={\begin{cases}{\frac {x^{n+1}}{n+1}}+C,&n\neq -1\\\ln \left|x\right|+C,&n=-1\end{cases}}}
∫ ∫ -->
d
x
a
2
+
x
2
=
1
a
arctg
x
a
+
C
=
− − -->
1
a
arcctg
x
a
+
C
{\displaystyle \int \!{dx \over {a^{2}+x^{2}}}={1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}+C=-{1 \over a}\,\operatorname {arcctg} \,{\frac {x}{a}}+C}
Сделаем замену
x
=
a
tg
-->
t
{\displaystyle x=a\operatorname {tg} t}
∫ ∫ -->
d
x
a
2
+
x
2
=
∫ ∫ -->
d
(
a
tg
-->
t
)
a
2
+
(
a
tg
-->
t
)
2
=
1
a
∫ ∫ -->
cos
2
-->
t
cos
2
-->
t
d
t
=
t
a
+
C
=
1
a
arctg
-->
x
a
+
C
.
{\displaystyle \int \!{dx \over {a^{2}+x^{2}}}=\int \!{d(a\operatorname {tg} t) \over a^{2}+(a\operatorname {tg} t)^{2}}={1 \over a}\int \!{\cos ^{2}t \over \cos ^{2}t}dt={t \over a}+C={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}
∫ ∫ -->
d
x
x
ln
-->
x
=
ln
-->
|
ln
-->
x
|
+
C
{\displaystyle \int {\frac {dx}{x\ln x}}=\ln |\ln x|+C}
∫ ∫ -->
log
b
-->
x
d
x
=
x
log
b
-->
x
− − -->
x
log
b
-->
e
+
C
=
x
ln
-->
x
− − -->
1
ln
-->
b
+
C
{\displaystyle \int \!\log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+C=x{\frac {\ln {x}-1}{\ln b}}+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}\,\operatorname {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}\pm a^{2}}}}=\ln \left|{x+{\sqrt {x^{2}\pm a^{2}}}}\right|+C}
∫ ∫ -->
x
2
+
a
d
x
=
1
2
(
x
x
2
+
a
+
a
ln
-->
|
x
+
x
2
+
a
|
)
+
C
{\displaystyle \int \!{\sqrt {x^{2}+a}}\,dx={1 \over 2}({x}{\sqrt {x^{2}+a}}+{a}\ln |x+{\sqrt {x^{2}+a}}|)+C}
Пусть
a
<
0
{\displaystyle a<0}
x
≥ ≥ -->
0
{\displaystyle x\geq 0}
гиперболическими функциями , сделаем замену
x
=
− − -->
a
ch
-->
t
,
t
≥ ≥ -->
0
{\displaystyle x={\sqrt {-a}}\operatorname {ch} t,t\geq 0}
∫ ∫ -->
x
2
+
a
d
x
=
∫ ∫ -->
(
− − -->
a
ch
-->
t
)
2
+
a
d
(
− − -->
a
ch
-->
t
)
=
− − -->
a
∫ ∫ -->
ch
2
-->
t
− − -->
1
sh
-->
t
d
t
=
− − -->
a
∫ ∫ -->
sh
2
-->
t
d
t
=
− − -->
a
∫ ∫ -->
ch
-->
2
t
− − -->
1
2
d
t
=
− − -->
a
2
(
sh
-->
2
t
2
− − -->
t
)
+
C
1
=
− − -->
a
2
(
sh
-->
t
ch
-->
t
− − -->
t
)
+
C
1
{\displaystyle {\begin{aligned}\int \!{\sqrt {x^{2}+a}}dx&=\int {\sqrt {({\sqrt {-a}}\operatorname {ch} t)^{2}+a}}d({\sqrt {-a}}\operatorname {ch} t)=-a\int {\sqrt {\operatorname {ch} ^{2}t-1}}\operatorname {sh} tdt\\&=-a\int \operatorname {sh} ^{2}tdt=-a\int {\operatorname {ch} 2t-1 \over 2}dt={-a \over 2}\left({\operatorname {sh} 2t \over 2}-t\right)+C_{1}\\&={-a \over 2}(\operatorname {sh} t\operatorname {ch} t-t)+C_{1}\end{aligned}}}
Но
sh
-->
t
=
ch
2
− − -->
1
=
x
2
− − -->
a
− − -->
1
=
x
2
+
a
− − -->
a
,
{\displaystyle \operatorname {sh} t={\sqrt {\operatorname {ch} ^{2}-1}}={\sqrt {{x^{2} \over -a}-1}}={{\sqrt {x^{2}+a}} \over {\sqrt {-a}}},}
sh
-->
t
ch
-->
t
=
x
x
2
+
a
− − -->
a
,
{\displaystyle \operatorname {sh} t\operatorname {ch} t=x{{\sqrt {x^{2}+a}} \over -a},}
e
t
=
sh
-->
t
+
ch
-->
t
=
x
+
x
2
+
a
− − -->
a
.
{\displaystyle e^{t}=\operatorname {sh} t+\operatorname {ch} t={x+{\sqrt {x^{2}+a}} \over {\sqrt {-a}}}.}
Поэтому
t
=
ln
-->
x
+
x
2
+
a
− − -->
a
.
{\displaystyle t=\ln {x+{\sqrt {x^{2}+a}} \over {\sqrt {-a}}}.}
Отсюда, включая логарифм знаменателя последней дроби в константу C, получаем
∫ ∫ -->
x
2
+
a
d
x
=
x
2
x
2
+
a
+
a
2
ln
-->
|
x
+
x
2
+
a
|
+
C
{\displaystyle \int \!{\sqrt {x^{2}+a}}\,dx={x \over 2}{\sqrt {x^{2}+a}}+{a \over 2}\ln |x+{\sqrt {x^{2}+a}}|+C}
Если
x
<
0
{\displaystyle x<0}
x
=
− − -->
t
,
t
>
0
{\displaystyle x=-t,t>0}
a
>
0
{\displaystyle a>0}
x
=
a
sh
-->
t
{\displaystyle x={\sqrt {a}}\operatorname {sh} t}
[1]
∫ ∫ -->
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}
∫ ∫ -->
tg
x
d
x
=
∫ ∫ -->
sin
-->
x
cos
-->
x
d
x
=
− − -->
∫ ∫ -->
d
(
cos
-->
x
)
cos
-->
x
=
− − -->
ln
-->
|
cos
-->
x
|
+
C
{\displaystyle \int \!\operatorname {tg} \,{x}\,dx=\int {\frac {\sin x}{\cos x}}dx=-\int {\frac {d(\cos x)}{\cos x}}=-\ln |\cos x|+C}
∫ ∫ -->
ctg
x
d
x
=
ln
-->
|
sin
-->
x
|
+
C
{\displaystyle \int \!\operatorname {ctg} \,{x}\,dx=\ln {\left|\sin {x}\right|}+C}
∫ ∫ -->
ctg
x
d
x
=
∫ ∫ -->
cos
-->
x
sin
-->
x
d
x
=
∫ ∫ -->
d
(
sin
-->
x
)
sin
-->
x
=
ln
-->
|
sin
-->
x
|
+
C
{\displaystyle \int \!\operatorname {ctg} \,{x}\,dx=\int {\frac {\cos x}{\sin x}}dx=\int {\frac {d(\sin x)}{\sin x}}=\ln |\sin x|+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
=
∫ ∫ -->
d
x
cos
2
-->
x
=
tg
x
+
C
{\displaystyle \int \!\sec ^{2}x\,dx=\int \!{dx \over \cos ^{2}x}=\operatorname {tg} \,x+C}
∫ ∫ -->
cosec
2
-->
x
d
x
=
∫ ∫ -->
d
x
sin
2
-->
x
=
− − -->
ctg
x
+
C
{\displaystyle \int \!\operatorname {cosec} ^{2}x\,dx=\int \!{dx \over \sin ^{2}x}=-\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}
∫ ∫ -->
sin
n
-->
x
d
x
=
− − -->
sin
n
− − -->
1
-->
x
cos
-->
x
n
+
n
− − -->
1
n
∫ ∫ -->
sin
n
− − -->
2
-->
x
d
x
,
n
∈ ∈ -->
N
,
n
⩾ ⩾ -->
2
{\displaystyle \int \!\sin ^{n}x\,dx=-{\frac {\sin ^{n-1}{x}\cos {x}}{n}}+{\frac {n-1}{n}}\int \!\sin ^{n-2}{x}\,dx,n\in \mathbb {N} ,n\geqslant 2}
∫ ∫ -->
cos
n
-->
x
d
x
=
cos
n
− − -->
1
-->
x
sin
-->
x
n
+
n
− − -->
1
n
∫ ∫ -->
cos
n
− − -->
2
-->
x
d
x
,
n
∈ ∈ -->
N
,
n
⩾ ⩾ -->
2
{\displaystyle \int \!\cos ^{n}x\,dx={\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \!\cos ^{n-2}{x}\,dx,n\in \mathbb {N} ,n\geqslant 2}
∫ ∫ -->
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}
∫ ∫ -->
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}
∫ ∫ -->
d
x
ch
2
x
=
th
x
+
C
{\displaystyle \int {\frac {dx}{\operatorname {ch} ^{2}\,x}}=\operatorname {th} \,x+C}
∫ ∫ -->
d
x
sh
2
x
=
− − -->
cth
x
+
C
{\displaystyle \int {\frac {dx}{\operatorname {sh} ^{2}\,x}}=-\operatorname {cth} \,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}
также
∫ ∫ -->
sech
x
d
x
=
2
arctg
(
e
x
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,(e^{x})+C}
также
∫ ∫ -->
sech
x
d
x
=
2
arctg
(
th
x
2
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C}
∫ ∫ -->
cth
x
d
x
=
ln
-->
|
sh
x
|
+
C
{\displaystyle \int \operatorname {cth} \,x\,dx=\ln |\operatorname {sh} \,x|+C}
Доказательство формулы
∫ ∫ -->
sech
x
d
x
=
arctg
-->
sh
x
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=\operatorname {arctg} \operatorname {sh} \,x+C}
∫ ∫ -->
sech
x
d
x
=
∫ ∫ -->
d
x
ch
-->
x
=
∫ ∫ -->
ch
-->
x
ch
2
-->
x
d
x
=
∫ ∫ -->
d
(
sh
-->
x
)
1
+
sh
2
-->
x
=
arctg
-->
sh
-->
x
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=\int {dx \over \operatorname {ch} x}=\int {\operatorname {ch} x \over \operatorname {ch} ^{2}x}dx=\int {d(\operatorname {sh} x) \over 1+\operatorname {sh} ^{2}x}=\operatorname {arctg} \operatorname {sh} x+C}
Доказательство формулы
∫ ∫ -->
sech
x
d
x
=
2
arctg
-->
(
e
x
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=2\operatorname {arctg} (e^{x})+C}
∫ ∫ -->
sech
x
d
x
=
∫ ∫ -->
d
x
ch
-->
x
=
2
∫ ∫ -->
d
x
e
x
+
e
− − -->
x
=
2
∫ ∫ -->
d
e
x
1
+
e
2
x
=
2
arctg
-->
(
e
x
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=\int {dx \over \operatorname {ch} x}=2\int {dx \over e^{x}+e^{-x}}=2\int {d{e^{x}} \over 1+e^{2x}}=2\operatorname {arctg} (e^{x})+C}
Доказательство формулы
∫ ∫ -->
sech
x
d
x
=
2
arctg
(
th
x
2
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C}
∫ ∫ -->
sech
x
d
x
=
∫ ∫ -->
1
ch
-->
x
d
x
=
∫ ∫ -->
d
x
sh
2
-->
x
2
+
ch
2
-->
x
2
=
2
∫ ∫ -->
d
(
x
2
)
ch
2
-->
x
2
(
1
+
th
2
-->
x
2
)
=
2
∫ ∫ -->
d
(
th
-->
x
2
)
1
+
th
2
-->
x
2
=
2
arctg
(
th
x
2
)
+
C
{\displaystyle {\begin{aligned}\int \operatorname {sech} \,x\,dx&=\int {1 \over \operatorname {ch} x}dx=\int {dx \over \operatorname {sh} ^{2}{x \over 2}+\operatorname {ch} ^{2}{x \over 2}}=2\int {d({x \over 2}) \over \operatorname {ch} ^{2}{x \over 2}(1+\operatorname {th} ^{2}{x \over 2})}\\&=2\int {d(\operatorname {th} {x \over 2}) \over 1+\operatorname {th} ^{2}{x \over 2}}=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C\end{aligned}}}
∫ ∫ -->
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 π -->
+
x
erf
-->
(
x
)
{\displaystyle \int \operatorname {erf} (x)\,dx={\frac {e^{-x^{2}}}{\sqrt {\pi }}}+x\operatorname {erf} (x)}
∫ ∫ -->
w
(
x
)
d
x
=
x
(
w
(
x
)
− − -->
1
+
1
w
(
x
)
)
{\displaystyle \int w(x)\,dx=x\left(w(x)-1+{\frac {1}{w(x)}}\right)}
Книги Таблицы интегралов Вычисление интегралов
Списки интегралов по типам функций
![{\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)
