Ռացիոնալ ֆունկցիաների ինտեգրալների ցանկ , ստորև ներկայացված են ռացիոնալ ֆունկցիաների ինտեգրալների ցանկերը։ Անորոշ ինտեգրալների համար ինտեգրման հաստատունը բաց է թողնված։
∫
(
a
x
+
b
)
n
d
x
=
{
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
,
n
≠
−
1
1
a
ln
|
a
x
+
b
|
,
n
=
−
1
{\displaystyle \int (ax+b)^{n}dx={\begin{cases}{\frac {(ax+b)^{n+1}}{a(n+1)}},&n\neq -1\\{\frac {1}{a}}\ln \left|ax+b\right|,&n=-1\end{cases}}}
∫
x
(
a
x
+
b
)
n
d
x
=
{
a
(
n
+
1
)
x
−
b
a
2
(
n
+
1
)
(
n
+
2
)
(
a
x
+
b
)
n
+
1
,
n
∉
{
−
1
,
−
2
}
x
a
−
b
a
2
ln
|
a
x
+
b
|
,
n
=
−
1
b
a
2
(
a
x
+
b
)
+
1
a
2
ln
|
a
x
+
b
|
,
n
=
−
2
{\displaystyle \int x(ax+b)^{n}dx={\begin{cases}{\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1},&n\not \in \{-1,-2\}\\{\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|,&n=-1\\{\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|,&n=-2\end{cases}}}
∫
x
(
a
x
+
b
)
n
d
x
=
a
(
1
−
n
)
x
−
b
a
2
(
n
−
1
)
(
n
−
2
)
(
a
x
+
b
)
n
−
1
,
n
∉
{
1
,
2
}
{\displaystyle \int {\frac {x}{(ax+b)^{n}}}dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}},\quad n\not \in \{1,2\}}
∫
d
x
x
2
n
+
1
=
∑
k
=
1
2
n
−
1
{
1
2
n
−
1
[
sin
(
(
2
k
−
1
)
π
2
n
)
arctan
[
(
x
−
cos
(
(
2
k
−
1
)
π
2
n
)
)
csc
(
(
2
k
−
1
)
π
2
n
)
]
]
−
1
2
n
[
cos
(
(
2
k
−
1
)
π
2
n
)
ln
|
x
2
−
2
x
cos
(
(
2
k
−
1
)
π
2
n
)
+
1
|
]
}
+
C
{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin \left({\frac {(2k-1)\pi }{2^{n}}}\right)\arctan \left[\left(x-\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right)\csc \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right]\right]-{\frac {1}{2^{n}}}\left[\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\ln \left|x^{2}-2x\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)+1\right|\right]\right\}+C}
∫
x
2
a
x
+
b
d
x
=
1
a
3
(
(
a
x
+
b
)
2
2
−
2
b
(
a
x
+
b
)
+
b
2
ln
|
a
x
+
b
|
)
{\displaystyle \int {\frac {x^{2}}{ax+b}}dx={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)}
∫
x
2
(
a
x
+
b
)
2
d
x
=
1
a
3
(
a
x
+
b
−
2
b
ln
|
a
x
+
b
|
−
b
2
a
x
+
b
)
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}dx={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)}
∫
x
2
(
a
x
+
b
)
3
d
x
=
1
a
3
(
ln
|
a
x
+
b
|
+
2
b
a
x
+
b
−
b
2
2
(
a
x
+
b
)
2
)
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)}
∫
x
2
(
a
x
+
b
)
n
d
x
=
1
a
3
(
−
1
(
n
−
3
)
(
a
x
+
b
)
n
−
3
+
2
b
(
n
−
2
)
(
a
x
+
b
)
n
−
2
−
b
2
(
n
−
1
)
(
a
x
+
b
)
n
−
1
)
,
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx={\frac {1}{a^{3}}}\left(-{\frac {1}{(n-3)(ax+b)^{n-3}}}+{\frac {2b}{(n-2)(ax+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right),}
n
∉
{
1
,
2
,
3
}
{\displaystyle \!n\not \in \{1,2,3\}}
∫
d
x
x
(
a
x
+
b
)
=
−
1
b
ln
|
a
x
+
b
x
|
{\displaystyle \int {\frac {dx}{x(ax+b)}}=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|}
∫
d
x
x
2
(
a
x
+
b
)
=
−
1
b
x
+
a
b
2
ln
|
a
x
+
b
x
|
{\displaystyle \int {\frac {dx}{x^{2}(ax+b)}}=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|}
∫
d
x
x
2
(
a
x
+
b
)
2
=
−
a
(
1
b
2
(
a
x
+
b
)
+
1
a
b
2
x
−
2
b
3
ln
|
a
x
+
b
x
|
)
{\displaystyle \int {\frac {dx}{x^{2}(ax+b)^{2}}}=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)}
∫
d
x
a
2
x
2
+
b
2
=
1
a
b
arctan
a
x
b
{\displaystyle \int {\frac {dx}{a^{2}x^{2}+b^{2}}}={\frac {1}{ab}}\arctan {\frac {ax}{b}}\,\!}
∫
d
x
(
x
2
+
a
2
)
2
=
x
2
a
2
(
x
2
+
a
2
)
+
1
2
a
3
arctan
x
a
{\displaystyle \int {\frac {dx}{(x^{2}+a^{2})^{2}}}={\frac {x}{2a^{2}(x^{2}+a^{2})}}+{\frac {1}{2a^{3}}}\arctan {\frac {x}{a}}}
∫
d
x
(
x
2
+
a
2
)
3
=
x
4
a
2
(
x
2
+
a
2
)
2
+
3
x
8
a
4
(
x
2
+
a
2
)
+
3
8
a
5
arctan
x
a
{\displaystyle \int {\frac {dx}{(x^{2}+a^{2})^{3}}}={\frac {x}{4a^{2}(x^{2}+a^{2})^{2}}}+{\frac {3x}{8a^{4}(x^{2}+a^{2})}}+{\frac {3}{8a^{5}}}\arctan {\frac {x}{a}}}
∫
d
x
x
2
−
a
2
=
−
1
a
a
r
t
a
n
h
x
a
=
1
2
a
ln
a
−
x
a
+
x
,
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {artanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}},}
|
x
|
<
|
a
|
{\displaystyle \!|x|<|a|}
∫
d
x
x
2
−
a
2
=
−
1
a
a
r
c
o
t
h
x
a
=
1
2
a
ln
x
−
a
x
+
a
,
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arcoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}},}
|
x
|
>
|
a
|
{\displaystyle \!|x|>|a|}
∫
d
x
a
x
2
+
b
x
+
c
=
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
,
{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}},}
4
a
c
−
b
2
>
0
{\displaystyle \!4ac-b^{2}>0}
∫
d
x
a
x
2
+
b
x
+
c
=
2
b
2
−
4
a
c
a
r
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
=
1
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
,
{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|,}
4
a
c
−
b
2
<
0
{\displaystyle \!4ac-b^{2}<0}
∫
d
x
a
x
2
+
b
x
+
c
=
−
2
2
a
x
+
b
(for
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}=-{\frac {2}{2ax+b}}\qquad {\mbox{(for }}4ac-b^{2}=0{\mbox{)}}}
∫
x
a
x
2
+
b
x
+
c
d
x
=
1
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
b
2
a
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \int {\frac {x}{ax^{2}+bx+c}}dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}}
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
+
2
a
n
−
b
m
a
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
,
{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}},}
4
a
c
−
b
2
>
0
{\displaystyle \!4ac-b^{2}>0}
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
2
a
n
−
b
m
a
b
2
−
4
a
c
a
r
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
,
{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}},}
4
a
c
−
b
2
<
0
{\displaystyle \!4ac-b^{2}<0}
∫
m
x
+
n
a
x
2
+
b
x
+
c
d
x
=
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
2
a
n
−
b
m
a
(
2
a
x
+
b
)
,
{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}},}
4
a
c
−
b
2
=
0
{\displaystyle \!4ac-b^{2}=0}
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
=
2
a
x
+
b
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
+
(
2
n
−
3
)
2
a
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
{\displaystyle \int {\frac {dx}{(ax^{2}+bx+c)^{n}}}={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {dx}{(ax^{2}+bx+c)^{n-1}}}\,\!}
∫
x
(
a
x
2
+
b
x
+
c
)
n
d
x
=
b
x
+
2
c
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
−
b
(
2
n
−
3
)
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
{\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx={\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {dx}{(ax^{2}+bx+c)^{n-1}}}\,\!}
∫
d
x
x
(
a
x
2
+
b
x
+
c
)
=
1
2
c
ln
|
x
2
a
x
2
+
b
x
+
c
|
−
b
2
c
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \int {\frac {dx}{x(ax^{2}+bx+c)}}={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {dx}{ax^{2}+bx+c}}}
Градштейн И. С. Рыжик И. М. Таблицы интегралов, сумм, рядов и произведений (4-е издание). М.։ Наука, 1963. ISBN 0-12-294757-6 // EqWorld
Двайт Г. Б. Таблицы интегралов СПб։ «Издательство и типография АО ВНИИГ им. Б. В. Веденеева», 1995.-176 с. ISBN 5-85529-029-8 .D. Zwillinger. CRC Standard Mathematical Tables and Formulae , 31st ed., 2002. ISBN 1-58488-291-3 .
M. Abramowitz and I. A. Stegun, eds. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables , 1964. ISBN 0-486-61272-4 Ինտեգրալների աղյուսակներ Խմբագրել Ինտեգրալների հաշվում Խմբագրել 
![{\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin \left({\frac {(2k-1)\pi }{2^{n}}}\right)\arctan \left[\left(x-\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right)\csc \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right]\right]-{\frac {1}{2^{n}}}\left[\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\ln \left|x^{2}-2x\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)+1\right|\right]\right\}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1462ab9faa5cae35bc52311fb0c84c7612541602)
