Seznam integralov racionalnih funkcij
Naslednji seznam vsebuje integrale racionalnih funkcij .
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
(za
n
≠
−
1
)
{\displaystyle \int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\text{(za }}n\neq -1{\mbox{)}}\,\!}
Cavalierijev obrazec za kvadraturo )
∫
c
a
x
+
b
d
x
=
c
a
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {c}{ax+b}}\,dx={\frac {c}{a}}\ln \left|ax+b\right|+C}
∫
x
(
a
x
+
b
)
n
d
x
=
a
(
n
+
1
)
x
−
b
a
2
(
n
+
1
)
(
n
+
2
)
(
a
x
+
b
)
n
+
1
+
C
(za
n
∉
{
−
1
,
−
2
}
)
{\displaystyle \int x(ax+b)^{n}\,dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}+C\qquad {\text{(za }}n\not \in \{-1,-2\}{\mbox{)}}}
∫
x
a
x
+
b
d
x
=
x
a
−
b
a
2
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {x}{ax+b}}\,dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|+C}
∫
x
(
a
x
+
b
)
2
d
x
=
b
a
2
(
a
x
+
b
)
+
1
a
2
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {x}{(ax+b)^{2}}}\,dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|+C}
∫
x
(
a
x
+
b
)
n
d
x
=
a
(
1
−
n
)
x
−
b
a
2
(
n
−
1
)
(
n
−
2
)
(
a
x
+
b
)
n
−
1
+
C
(za
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}}}+C\qquad {\text{(za }}n\not \in \{1,2\}{\mbox{)}}}
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
|
f
(
x
)
|
+
C
{\displaystyle \int {\frac {f'(x)}{f(x)}}\,dx=\ln \left|f(x)\right|+C}
∫
x
2
a
x
+
b
d
x
=
b
2
ln
(
|
a
x
+
b
|
)
a
3
+
a
x
2
−
2
b
x
2
a
2
+
C
{\displaystyle \int {\frac {x^{2}}{ax+b}}\,dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3}}}+{\frac {ax^{2}-2bx}{2a^{2}}}+C}
∫
x
2
(
a
x
+
b
)
2
d
x
=
1
a
3
(
a
x
−
2
b
ln
|
a
x
+
b
|
−
b
2
a
x
+
b
)
+
C
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}\,dx={\frac {1}{a^{3}}}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)+C}
∫
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
)
+
C
{\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)+C}
∫
x
2
(
a
x
+
b
)
n
d
x
=
1
a
3
(
−
(
a
x
+
b
)
3
−
n
(
n
−
3
)
+
2
b
(
a
x
+
b
)
2
−
n
(
n
−
2
)
−
b
2
(
a
x
+
b
)
1
−
n
(
n
−
1
)
)
+
C
(za
n
∉
{
1
,
2
,
3
}
)
{\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}\,dx={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(ax+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)+C\qquad {\text{(za }}n\not \in \{1,2,3\}{\mbox{)}}}
∫
1
x
(
a
x
+
b
)
d
x
=
−
1
b
ln
|
a
x
+
b
x
|
+
C
{\displaystyle \int {\frac {1}{x(ax+b)}}\,dx=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C}
∫
1
x
2
(
a
x
+
b
)
d
x
=
−
1
b
x
+
a
b
2
ln
|
a
x
+
b
x
|
+
C
{\displaystyle \int {\frac {1}{x^{2}(ax+b)}}\,dx=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C}
∫
1
x
2
(
a
x
+
b
)
2
d
x
=
−
a
(
1
b
2
(
a
x
+
b
)
+
1
a
b
2
x
−
2
b
3
ln
|
a
x
+
b
x
|
)
+
C
{\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}\,dx=-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)+C}
∫
1
x
2
+
a
2
d
x
=
1
a
arctan
x
a
+
C
{\displaystyle \int {\frac {1}{x^{2}+a^{2}}}\,dx={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!+C}
∫
1
x
2
−
a
2
d
x
=
{
−
1
a
a
r
c
t
a
n
h
x
a
=
1
2
a
ln
a
−
x
a
+
x
+
C
(za
|
x
|
<
|
a
|
)
−
1
a
a
r
c
c
o
t
h
x
a
=
1
2
a
ln
x
−
a
x
+
a
+
C
(za
|
x
|
>
|
a
|
)
{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\begin{cases}\displaystyle -{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(za }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(za }}|x|>|a|{\mbox{)}}\end{cases}}}
∫
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}
Vsako racionalno funkcijo lahko integriramo tako, da uporabimo zgornje enačbe in delne ulomke
a
(
x
−
b
)
n
{\displaystyle {\frac {a}{(x-b)^{n}}}}
a
x
+
b
(
(
x
−
c
)
2
+
d
2
)
n
.
{\displaystyle {\frac {ax+b}{\left((x-c)^{2}+d^{2}\right)^{n}}}.}
x
m
(
a
x
2
+
b
x
+
c
)
n
{\displaystyle {\frac {x^{m}}{(a\,x^{2}+b\,x+c)^{n}}}}
uredi
Za
a
≠
0
:
{\displaystyle a\neq 0:}
∫
1
a
x
2
+
b
x
+
c
d
x
=
{
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
+
C
(za
4
a
c
−
b
2
>
0
)
−
2
b
2
−
4
a
c
a
r
c
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
+
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
|
+
C
(za
4
a
c
−
b
2
<
0
)
−
2
2
a
x
+
b
+
C
(za
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(za }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C&{\text{(za }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\text{(za }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}
∫
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
+
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}}+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
+
C
(za
4
a
c
−
b
2
>
0
)
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
2
a
n
−
b
m
a
b
2
−
4
a
c
a
r
c
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
+
C
(za
4
a
c
−
b
2
<
0
)
m
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
2
a
n
−
b
m
a
(
2
a
x
+
b
)
+
C
(za
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}\,dx={\begin{cases}\displaystyle {\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}}}}+C&{\text{(za }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(za }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C&{\text{(za }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}
∫
1
(
a
x
2
+
b
x
+
c
)
n
d
x
=
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
)
∫
1
(
a
x
2
+
b
x
+
c
)
n
−
1
d
x
+
C
{\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,dx={\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 {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}
∫
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
)
∫
1
(
a
x
2
+
b
x
+
c
)
n
−
1
d
x
+
C
{\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 {1}{(ax^{2}+bx+c)^{n-1}}}\,dx+C}
∫
1
x
(
a
x
2
+
b
x
+
c
)
d
x
=
1
2
c
ln
|
x
2
a
x
2
+
b
x
+
c
|
−
b
2
c
∫
1
a
x
2
+
b
x
+
c
d
x
+
C
{\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}\,dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}\,dx+C}
x
m
(
a
+
b
x
n
)
p
{\displaystyle x^{m}\left(a+b\,x^{n}\right)^{p}}
uredi
Integrand, ki ga dobimo ima enako obliko kot prvotni integrand tako, da lahko ponavljamo nižanje potenc tako, da nižamo potenci m p To zmanjšanje potenc lahko uporabimo za integrande, ki imajo celoštevilčne ali ulomljene potence.
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
+
1
(
a
+
b
x
n
)
p
m
+
n
p
+
1
+
a
n
p
m
+
n
p
+
1
∫
x
m
(
a
+
b
x
n
)
p
−
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+n\,p+1}}\,+\,{\frac {a\,n\,p}{m+n\,p+1}}\int x^{m}\left(a+b\,x^{n}\right)^{p-1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
−
x
m
+
1
(
a
+
b
x
n
)
p
+
1
a
n
(
p
+
1
)
+
m
+
n
(
p
+
1
)
+
1
a
n
(
p
+
1
)
∫
x
m
(
a
+
b
x
n
)
p
+
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx=-{\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a\,n(p+1)}}\,+\,{\frac {m+n(p+1)+1}{a\,n(p+1)}}\int x^{m}\left(a+b\,x^{n}\right)^{p+1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
+
1
(
a
+
b
x
n
)
p
m
+
1
−
b
n
p
m
+
1
∫
x
m
+
n
(
a
+
b
x
n
)
p
−
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p}}{m+1}}\,-\,{\frac {b\,n\,p}{m+1}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p-1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
−
n
+
1
(
a
+
b
x
n
)
p
+
1
b
n
(
p
+
1
)
−
m
−
n
+
1
b
n
(
p
+
1
)
∫
x
m
−
n
(
a
+
b
x
n
)
p
+
1
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b\,n(p+1)}}\,-\,{\frac {m-n+1}{b\,n(p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p+1}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
−
n
+
1
(
a
+
b
x
n
)
p
+
1
b
(
m
+
n
p
+
1
)
−
a
(
m
−
n
+
1
)
b
(
m
+
n
p
+
1
)
∫
x
m
−
n
(
a
+
b
x
n
)
p
d
x
{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}}{b(m+n\,p+1)}}\,-\,{\frac {a(m-n+1)}{b(m+n\,p+1)}}\int x^{m-n}\left(a+b\,x^{n}\right)^{p}dx}
∫
x
m
(
a
+
b
x
n
)
p
d
x
=
x
m
+
1
(
a
+
b
x
n
)
p
+
1
a
(
m
+
1
)
−
b
(
m
+
n
(
p
+
1
)
+
1
)
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{\displaystyle \int x^{m}\left(a+b\,x^{n}\right)^{p}dx={\frac {x^{m+1}\left(a+b\,x^{n}\right)^{p+1}}{a(m+1)}}\,-\,{\frac {b(m+n(p+1)+1)}{a(m+1)}}\int x^{m+n}\left(a+b\,x^{n}\right)^{p}dx}
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{\displaystyle x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}}
uredi
Nastali integrandi imajo enako obliko kot prvotni integrandi, to pa pomeni, da se zniževanje potence lahko ponavlja z zmanjševanjem potenc m p q To zmanjševanje potenc se lahko uporabi za integrande, ki imajo celoštevilčne ali ulomljene eksponente. Posebni primer takšnega zmanjševanja potenc se lahko uporabi za integrande v obliki
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{\displaystyle \left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}}
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{\displaystyle x^{m}\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a\,b\,n(p+1)}}\,+\,{\frac {1}{a\,b\,n(p+1)}}\,\cdot }
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{\displaystyle \int x^{m}\left(c(A\,b\,n(p+1)+(A\,b-a\,B)(m+1))+d(A\,b\,n(p+1)+(A\,b-a\,B)(m+n\,q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q-1}dx}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{b(m+n(p+q+1)+1)}}\,+\,{\frac {1}{b(m+n(p+q+1)+1)}}\,\cdot }
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{\displaystyle \int x^{m}\left(c((A\,b-a\,B)(1+m)+A\,b\,n(1+p+q))+(d(A\,b-a\,B)(1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n(1+p+q))\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx=-{\frac {(A\,b-a\,B)x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,n(b\,c-a\,d)(p+1)}}\,+\,{\frac {1}{a\,n(b\,c-a\,d)(p+1)}}\,\cdot }
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{\displaystyle \int x^{m}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c-a\,d)(p+1)+d(A\,b-a\,B)(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {B\,x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,d(m+n(p+q+1)+1)}}\,-\,{\frac {1}{b\,d(m+n(p+q+1)+1)}}\,\cdot }
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{\displaystyle \int x^{m-n}\left(a\,B\,c(m-n+1)+(a\,B\,d(m+n\,q+1)-b(-B\,c(m+n\,p+1)+A\,d(m+n(p+q+1)+1)))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{a\,c(m+1)}}\,+\,{\frac {1}{a\,c(m+1)}}\,\cdot }
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{\displaystyle \int x^{m+n}\left(a\,B\,c(m+1)-A(b\,c+a\,d)(m+n+1)-A\,n(b\,c\,p+a\,d\,q)-A\,b\,d(m+n(p+q+2)+1)x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {A\,x^{m+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}}{a(m+1)}}\,-\,{\frac {1}{a(m+1)}}\,\cdot }
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{\displaystyle \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n(b\,c(p+1)+a\,d\,q)+d((A\,b-a\,B)(m+1)+A\,b\,n(p+q+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q-1}dx}
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{\displaystyle \int x^{m}\left(A+B\,x^{n}\right)\left(a+b\,x^{n}\right)^{p}\left(c+d\,x^{n}\right)^{q}dx={\frac {(A\,b-a\,B)x^{m-n+1}\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q+1}}{b\,n(b\,c-a\,d)(p+1)}}\,-\,{\frac {1}{b\,n(b\,c-a\,d)(p+1)}}\,\cdot }
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{\displaystyle \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1))x^{n}\right)\left(a+b\,x^{n}\right)^{p+1}\left(c+d\,x^{n}\right)^{q}dx}
![{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\begin{cases}\displaystyle -{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\text{(za }}|x|<|a|{\mbox{)}}\\[12pt]\displaystyle -{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\text{(za }}|x|>|a|{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8ad2f2e72c97bb788e3ab057fe66e00c078e48a)
![{\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)
![{\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\text{(za }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C&{\text{(za }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\text{(za }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c22d96d88878a10e8c1d667d2a5da4258630609)
![{\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}\,dx={\begin{cases}\displaystyle {\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}}}}+C&{\text{(za }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\text{(za }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle {\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C&{\text{(za }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42dfc5c00947fd17b8d5fa9b66c7e5dd5f11d428)
