Ниже приведён список интегралов (первообразных функций) от иррациональных функций . В списке везде опущена аддитивная константа интегрирования.
Везде ниже:
r
=
a
2
+
x
2
{\displaystyle r={\sqrt {a^{2}+x^{2}}}}
∫
r
d
x
=
1
2
(
x
r
+
a
2
ln
(
x
+
r
)
)
{\displaystyle \int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left({x+r}\right)\right)}
∫
r
3
d
x
=
1
4
x
r
3
+
1
8
3
a
2
x
r
+
3
8
a
4
ln
(
x
+
r
a
)
{\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {1}{8}}3a^{2}xr+{\frac {3}{8}}a^{4}\ln \left({\frac {x+r}{a}}\right)}
∫
r
5
d
x
=
1
6
x
r
5
+
5
24
a
2
x
r
3
+
5
16
a
4
x
r
+
5
16
a
6
ln
(
x
+
r
a
)
{\displaystyle \int r^{5}\;dx={\frac {1}{6}}xr^{5}+{\frac {5}{24}}a^{2}xr^{3}+{\frac {5}{16}}a^{4}xr+{\frac {5}{16}}a^{6}\ln \left({\frac {x+r}{a}}\right)}
∫
x
r
2
n
+
1
d
x
=
r
2
n
+
3
2
n
+
3
{\displaystyle \int xr^{2n+1}\;dx={\frac {r^{2n+3}}{2n+3}}}
∫
x
2
r
d
x
=
x
r
3
4
−
a
2
x
r
8
−
a
4
8
ln
(
x
+
r
a
)
{\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left({\frac {x+r}{a}}\right)}
∫
x
2
r
3
d
x
=
x
r
5
6
−
a
2
x
r
3
24
−
a
4
x
r
16
−
a
6
16
ln
(
x
+
r
a
)
{\displaystyle \int x^{2}r^{3}\;dx={\frac {xr^{5}}{6}}-{\frac {a^{2}xr^{3}}{24}}-{\frac {a^{4}xr}{16}}-{\frac {a^{6}}{16}}\ln \left({\frac {x+r}{a}}\right)}
∫
x
3
r
d
x
=
r
5
5
−
a
2
r
3
3
{\displaystyle \int x^{3}r\;dx={\frac {r^{5}}{5}}-{\frac {a^{2}r^{3}}{3}}}
∫
x
3
r
3
d
x
=
r
7
7
−
a
2
r
5
5
{\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{7}}{7}}-{\frac {a^{2}r^{5}}{5}}}
∫
x
3
r
2
n
+
1
d
x
=
r
2
n
+
5
2
n
+
5
−
a
3
r
2
n
+
3
2
n
+
3
{\displaystyle \int x^{3}r^{2n+1}\;dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}}
∫
x
4
r
d
x
=
x
3
r
3
6
−
a
2
x
r
3
8
+
a
4
x
r
16
+
a
6
16
ln
(
x
+
r
a
)
{\displaystyle \int x^{4}r\;dx={\frac {x^{3}r^{3}}{6}}-{\frac {a^{2}xr^{3}}{8}}+{\frac {a^{4}xr}{16}}+{\frac {a^{6}}{16}}\ln \left({\frac {x+r}{a}}\right)}
∫
x
4
r
3
d
x
=
x
3
r
5
8
−
a
2
x
r
5
16
+
a
4
x
r
3
64
+
3
a
6
x
r
128
+
3
a
8
128
ln
(
x
+
r
a
)
{\displaystyle \int x^{4}r^{3}\;dx={\frac {x^{3}r^{5}}{8}}-{\frac {a^{2}xr^{5}}{16}}+{\frac {a^{4}xr^{3}}{64}}+{\frac {3a^{6}xr}{128}}+{\frac {3a^{8}}{128}}\ln \left({\frac {x+r}{a}}\right)}
∫
x
5
r
d
x
=
r
7
7
−
2
a
2
r
5
5
+
a
4
r
3
3
{\displaystyle \int x^{5}r\;dx={\frac {r^{7}}{7}}-{\frac {2a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}}
∫
x
5
r
3
d
x
=
r
9
9
−
2
a
2
r
7
7
+
a
4
r
5
5
{\displaystyle \int x^{5}r^{3}\;dx={\frac {r^{9}}{9}}-{\frac {2a^{2}r^{7}}{7}}+{\frac {a^{4}r^{5}}{5}}}
∫
x
5
r
2
n
+
1
d
x
=
r
2
n
+
7
2
n
+
7
−
2
a
2
r
2
n
+
5
2
n
+
5
+
a
4
r
2
n
+
3
2
n
+
3
{\displaystyle \int x^{5}r^{2n+1}\;dx={\frac {r^{2n+7}}{2n+7}}-{\frac {2a^{2}r^{2n+5}}{2n+5}}+{\frac {a^{4}r^{2n+3}}{2n+3}}}
∫
r
d
x
x
=
r
−
a
ln
|
a
+
r
x
|
=
r
−
a
arsh
a
x
{\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\operatorname {arsh} {\frac {a}{x}}}
∫
r
3
d
x
x
=
r
3
3
+
a
2
r
−
a
3
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {r^{3}\;dx}{x}}={\frac {r^{3}}{3}}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|}
∫
r
5
d
x
x
=
r
5
5
+
a
2
r
3
3
+
a
4
r
−
a
5
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {r^{5}\;dx}{x}}={\frac {r^{5}}{5}}+{\frac {a^{2}r^{3}}{3}}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}}\right|}
∫
r
7
d
x
x
=
r
7
7
+
a
2
r
5
5
+
a
4
r
3
3
+
a
6
r
−
a
7
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {r^{7}\;dx}{x}}={\frac {r^{7}}{7}}+{\frac {a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}+a^{6}r-a^{7}\ln \left|{\frac {a+r}{x}}\right|}
∫
d
x
r
=
arsh
x
a
=
ln
|
x
+
r
|
{\displaystyle \int {\frac {dx}{r}}=\operatorname {arsh} {\frac {x}{a}}=\ln \left|x+r\right|}
∫
x
d
x
r
=
r
{\displaystyle \int {\frac {x\,dx}{r}}=r}
∫
x
2
d
x
r
=
x
2
r
−
a
2
2
arsh
x
a
=
x
2
r
−
a
2
2
ln
|
x
+
r
|
{\displaystyle \int {\frac {x^{2}\;dx}{r}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\,\operatorname {arsh} {\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left|x+r\right|}
∫
d
x
x
r
=
−
1
a
arsh
a
x
=
−
1
a
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\operatorname {arsh} {\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|}
Везде ниже:
s
=
x
2
−
a
2
{\displaystyle s={\sqrt {x^{2}-a^{2}}}}
Принято
x
2
>
a
2
{\displaystyle x^{2}>a^{2}}
x
2
<
a
2
{\displaystyle x^{2}<a^{2}}
∫
s
d
x
=
1
2
(
x
s
−
a
2
ln
(
x
+
s
)
)
{\displaystyle \int s\;dx={\frac {1}{2}}\left(xs-a^{2}\ln \left(x+s\right)\right)}
∫
x
s
d
x
=
−
1
3
s
3
{\displaystyle \int xs\;dx=-{\frac {1}{3}}s^{3}}
∫
s
d
x
x
=
s
−
a
cos
−
1
|
a
x
|
{\displaystyle \int {\frac {s\;dx}{x}}=s-a\cos ^{-1}\left|{\frac {a}{x}}\right|}
∫
d
x
s
=
∫
d
x
x
2
−
a
2
=
ln
|
x
+
s
|
{\displaystyle \int {\frac {dx}{s}}=\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\ln \left|x+s\right|}
Заметим, что
ln
|
x
+
s
a
|
=
s
g
n
(
x
)
arch
|
x
a
|
=
1
2
ln
(
x
+
s
x
−
s
)
{\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\mathrm {sgn} (x)\operatorname {arch} \left|{\frac {x}{a}}\right|={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)}
arch
|
x
a
|
{\displaystyle \operatorname {arch} \left|{\frac {x}{a}}\right|}
∫
x
d
x
s
=
s
{\displaystyle \int {\frac {x\;dx}{s}}=s}
∫
x
d
x
s
3
=
−
1
s
{\displaystyle \int {\frac {x\;dx}{s^{3}}}=-{\frac {1}{s}}}
∫
x
d
x
s
5
=
−
1
3
s
3
{\displaystyle \int {\frac {x\;dx}{s^{5}}}=-{\frac {1}{3s^{3}}}}
∫
x
d
x
s
7
=
−
1
5
s
5
{\displaystyle \int {\frac {x\;dx}{s^{7}}}=-{\frac {1}{5s^{5}}}}
∫
x
d
x
s
2
n
+
1
=
−
1
(
2
n
−
1
)
s
2
n
−
1
{\displaystyle \int {\frac {x\;dx}{s^{2n+1}}}=-{\frac {1}{(2n-1)s^{2n-1}}}}
∫
x
2
m
d
x
s
2
n
+
1
=
−
1
2
n
−
1
x
2
m
−
1
s
2
n
−
1
+
2
m
−
1
2
n
−
1
∫
x
2
m
−
2
d
x
s
2
n
−
1
{\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=-{\frac {1}{2n-1}}{\frac {x^{2m-1}}{s^{2n-1}}}+{\frac {2m-1}{2n-1}}\int {\frac {x^{2m-2}\;dx}{s^{2n-1}}}}
∫
x
2
d
x
s
=
x
s
2
+
a
2
2
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{2}\;dx}{s}}={\frac {xs}{2}}+{\frac {a^{2}}{2}}\ln \left|{\frac {x+s}{a}}\right|}
∫
x
2
d
x
s
3
=
−
x
s
+
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{2}\;dx}{s^{3}}}=-{\frac {x}{s}}+\ln \left|{\frac {x+s}{a}}\right|}
∫
x
4
d
x
s
=
x
3
s
4
+
3
8
a
2
x
s
+
3
8
a
4
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{4}\;dx}{s}}={\frac {x^{3}s}{4}}+{\frac {3}{8}}a^{2}xs+{\frac {3}{8}}a^{4}\ln \left|{\frac {x+s}{a}}\right|}
∫
x
4
d
x
s
3
=
x
s
2
−
a
2
x
s
+
3
2
a
2
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{4}\;dx}{s^{3}}}={\frac {xs}{2}}-{\frac {a^{2}x}{s}}+{\frac {3}{2}}a^{2}\ln \left|{\frac {x+s}{a}}\right|}
∫
x
4
d
x
s
5
=
−
x
s
−
1
3
x
3
s
3
+
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{4}\;dx}{s^{5}}}=-{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}+\ln \left|{\frac {x+s}{a}}\right|}
∫
x
2
m
d
x
s
2
n
+
1
=
(
−
1
)
n
−
m
1
a
2
(
n
−
m
)
∑
i
=
0
n
−
m
−
1
1
2
(
m
+
i
)
+
1
(
n
−
m
−
1
i
)
x
2
(
m
+
i
)
+
1
s
2
(
m
+
i
)
+
1
,
{\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=(-1)^{n-m}{\frac {1}{a^{2(n-m)}}}\sum _{i=0}^{n-m-1}{\frac {1}{2(m+i)+1}}{n-m-1 \choose i}{\frac {x^{2(m+i)+1}}{s^{2(m+i)+1}}},}
n
>
m
≥
0
{\displaystyle n>m\geq 0}
∫
d
x
s
3
=
−
1
a
2
x
s
{\displaystyle \int {\frac {dx}{s^{3}}}=-{\frac {1}{a^{2}}}{\frac {x}{s}}}
∫
d
x
s
5
=
1
a
4
[
x
s
−
1
3
x
3
s
3
]
{\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}
∫
d
x
s
7
=
−
1
a
6
[
x
s
−
2
3
x
3
s
3
+
1
5
x
5
s
5
]
{\displaystyle \int {\frac {dx}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\frac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}
∫
d
x
s
9
=
1
a
8
[
x
s
−
3
3
x
3
s
3
+
3
5
x
5
s
5
−
1
7
x
7
s
7
]
{\displaystyle \int {\frac {dx}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\frac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}
∫
x
2
d
x
s
5
=
−
1
a
2
x
3
3
s
3
{\displaystyle \int {\frac {x^{2}\;dx}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}}
∫
x
2
d
x
s
7
=
1
a
4
[
1
3
x
3
s
3
−
1
5
x
5
s
5
]
{\displaystyle \int {\frac {x^{2}\;dx}{s^{7}}}={\frac {1}{a^{4}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}
∫
x
2
d
x
s
9
=
−
1
a
6
[
1
3
x
3
s
3
−
2
5
x
5
s
5
+
1
7
x
7
s
7
]
{\displaystyle \int {\frac {x^{2}\;dx}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}
Везде ниже:
t
=
a
2
−
x
2
(
|
x
|
⩽
|
a
|
)
{\displaystyle t={\sqrt {a^{2}-x^{2}}}\qquad {\mbox{(}}|x|\leqslant |a|{\mbox{)}}}
∫
t
d
x
=
1
2
(
x
t
+
a
2
arcsin
x
a
)
=
1
2
(
x
t
−
a
2
arccos
x
a
)
{\displaystyle \int t\;dx={\frac {1}{2}}\left(xt+a^{2}\arcsin {\frac {x}{a}}\right)={\frac {1}{2}}\left(xt-a^{2}\arccos {\frac {x}{a}}\right)}
∫
x
t
d
x
=
−
1
3
t
3
{\displaystyle \int xt\;dx=-{\frac {1}{3}}t^{3}}
∫
t
d
x
x
=
t
−
a
ln
|
a
+
t
x
|
{\displaystyle \int {\frac {t\;dx}{x}}=t-a\ln \left|{\frac {a+t}{x}}\right|}
∫
d
x
t
=
arcsin
x
a
{\displaystyle \int {\frac {dx}{t}}=\arcsin {\frac {x}{a}}}
∫
x
d
x
t
=
−
t
{\displaystyle \int {\frac {x\;dx}{t}}=-t}
∫
x
2
d
x
t
=
−
x
2
t
+
a
2
2
arcsin
x
a
{\displaystyle \int {\frac {x^{2}\;dx}{t}}=-{\frac {x}{2}}t+{\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}}
∫
t
d
x
=
1
2
(
x
t
−
sgn
x
arch
|
x
a
|
)
{\displaystyle \int t\;dx={\frac {1}{2}}\left(xt-\operatorname {sgn} x\,\operatorname {arch} \left|{\frac {x}{a}}\right|\right)}
Здесь обозначено:
R
=
a
x
2
+
b
x
+
c
{\displaystyle R=ax^{2}+bx+c}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
ln
|
2
a
R
+
2
a
x
+
b
|
(
a
>
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {aR}}+2ax+b\right|\qquad {\mbox{( }}a>0{\mbox{)}}}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
arsh
2
a
x
+
b
4
a
c
−
b
2
(
a
>
0
,
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\,\operatorname {arsh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{( }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
ln
|
2
a
x
+
b
|
(
a
>
0
,
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{( }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}
∫
d
x
a
x
2
+
b
x
+
c
=
−
1
−
a
arcsin
2
a
x
+
b
b
2
−
4
a
c
(
a
<
0
,
4
a
c
−
b
2
<
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{( }}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{)}}}
∫
d
x
(
a
x
2
+
b
x
+
c
)
3
=
4
a
x
+
2
b
(
4
a
c
−
b
2
)
R
{\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{3}}}}={\frac {4ax+2b}{(4ac-b^{2}){\sqrt {R}}}}}
∫
d
x
(
a
x
2
+
b
x
+
c
)
5
=
4
a
x
+
2
b
3
(
4
a
c
−
b
2
)
R
(
1
R
+
8
a
4
a
c
−
b
2
)
{\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{5}}}}={\frac {4ax+2b}{3(4ac-b^{2}){\sqrt {R}}}}\left({\frac {1}{R}}+{\frac {8a}{4ac-b^{2}}}\right)}
∫
d
x
(
a
x
2
+
b
x
+
c
)
2
n
+
1
=
4
a
x
+
2
b
(
2
n
−
1
)
(
4
a
c
−
b
2
)
R
(
2
n
−
1
)
/
2
+
8
a
(
n
−
1
)
(
2
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
R
(
2
n
−
1
)
/
2
{\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}={\frac {4ax+2b}{(2n-1)(4ac-b^{2})R^{(2n-1)/2}}}+{\frac {8a(n-1)}{(2n-1)(4ac-b^{2})}}\int {\frac {dx}{R^{(2n-1)/2}}}}
∫
x
d
x
a
x
2
+
b
x
+
c
=
R
a
−
b
2
a
∫
d
x
R
{\displaystyle \int {\frac {x\;dx}{\sqrt {ax^{2}+bx+c}}}={\frac {\sqrt {R}}{a}}-{\frac {b}{2a}}\int {\frac {dx}{\sqrt {R}}}}
∫
x
d
x
(
a
x
2
+
b
x
+
c
)
3
=
−
2
b
x
+
4
c
(
4
a
c
−
b
2
)
R
{\displaystyle \int {\frac {x\;dx}{\sqrt {(ax^{2}+bx+c)^{3}}}}=-{\frac {2bx+4c}{(4ac-b^{2}){\sqrt {R}}}}}
∫
x
d
x
(
a
x
2
+
b
x
+
c
)
2
n
+
1
=
−
1
(
2
n
−
1
)
a
R
(
2
n
−
1
)
/
2
−
b
2
a
∫
d
x
R
(
2
n
+
1
)
/
2
{\displaystyle \int {\frac {x\;dx}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}=-{\frac {1}{(2n-1)aR^{(2n-1)/2}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{(2n+1)/2}}}}
∫
d
x
x
a
x
2
+
b
x
+
c
=
−
1
c
ln
(
2
c
R
+
b
x
+
2
c
x
)
{\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {cR}}+bx+2c}{x}}\right)}
∫
d
x
x
a
x
2
+
b
x
+
c
=
−
1
c
arsh
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
{\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}
∫
a
x
2
+
b
x
+
c
d
x
=
(
x
2
+
b
4
a
)
a
x
2
+
b
x
+
c
+
(
c
2
−
b
2
8
a
)
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \int {\sqrt {ax^{2}+bx+c}}\,dx=({\frac {x}{2}}+{\frac {b}{4a}}){\sqrt {ax^{2}+bx+c}}+({\frac {c}{2}}-{\frac {b^{2}}{8a}})\int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}\qquad }
∫
d
x
x
a
x
+
b
=
−
2
b
arth
a
x
+
b
b
{\displaystyle \int {\frac {dx}{x{\sqrt {ax+b}}}}\,=\,{\frac {-2}{\sqrt {b}}}\operatorname {arth} {\sqrt {\frac {ax+b}{b}}}}
∫
a
x
+
b
x
d
x
=
2
(
a
x
+
b
−
b
arth
a
x
+
b
b
)
{\displaystyle \int {\frac {\sqrt {ax+b}}{x}}\,dx\;=\;2\left({\sqrt {ax+b}}-{\sqrt {b}}\operatorname {arth} {\sqrt {\frac {ax+b}{b}}}\right)}
∫
x
n
a
x
+
b
d
x
=
2
a
(
2
n
+
1
)
(
x
n
a
x
+
b
−
b
n
∫
x
n
−
1
a
x
+
b
d
x
)
{\displaystyle \int {\frac {x^{n}}{\sqrt {ax+b}}}\,dx\;=\;{\frac {2}{a(2n+1)}}\left(x^{n}{\sqrt {ax+b}}-bn\int {\frac {x^{n-1}}{\sqrt {ax+b}}}\,dx\right)}
∫
x
n
a
x
+
b
d
x
=
2
2
n
+
1
(
x
n
+
1
a
x
+
b
+
b
x
n
a
x
+
b
−
n
b
∫
x
n
−
1
a
x
+
b
d
x
)
{\displaystyle \int x^{n}{\sqrt {ax+b}}\,dx\;=\;{\frac {2}{2n+1}}\left(x^{n+1}{\sqrt {ax+b}}+bx^{n}{\sqrt {ax+b}}-nb\int x^{n-1}{\sqrt {ax+b}}\,dx\right)}
Книги Таблицы интегралов Вычисление интегралов
Списки интегралов по типам функций
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![{\displaystyle \int {\frac {x^{2}\;dx}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/239ff6c41a3342440c712b9f0c4940e8e6a000d2)
