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From Wikiversity
∫
cot
a
x
d
x
=
1
a
ln
|
sin
a
x
|
+
C
{\displaystyle \int \cot ax\;dx={\frac {1}{a}}\ln |\sin ax|+C\,\!}
∫
cot
n
a
x
d
x
=
−
1
a
(
n
−
1
)
cot
n
−
1
a
x
−
∫
cot
n
−
2
a
x
d
x
(for
n
≠
1
)
{\displaystyle \int \cot ^{n}ax\;dx=-{\frac {1}{a(n-1)}}\cot ^{n-1}ax-\int \cot ^{n-2}ax\;dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
1
+
cot
a
x
=
∫
tan
a
x
d
x
tan
a
x
+
1
{\displaystyle \int {\frac {dx}{1+\cot ax}}=\int {\frac {\tan ax\;dx}{\tan ax+1}}\,\!}
∫
d
x
1
−
cot
a
x
=
∫
tan
a
x
d
x
tan
a
x
−
1
{\displaystyle \int {\frac {dx}{1-\cot ax}}=\int {\frac {\tan ax\;dx}{\tan ax-1}}\,\!}
∫
d
x
cos
a
x
±
sin
a
x
=
1
a
2
ln
|
tan
(
a
x
2
±
π
8
)
|
+
C
{\displaystyle \int {\frac {dx}{\cos ax\pm \sin ax}}={\frac {1}{a{\sqrt {2}}}}\ln \left|\tan \left({\frac {ax}{2}}\pm {\frac {\pi }{8}}\right)\right|+C}
∫
d
x
(
cos
a
x
±
sin
a
x
)
2
=
1
2
a
tan
(
a
x
∓
π
4
)
+
C
{\displaystyle \int {\frac {dx}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C}
∫
d
x
(
cos
x
+
sin
x
)
n
=
1
n
−
1
(
sin
x
−
cos
x
(
cos
x
+
sin
x
)
n
−
1
−
2
(
n
−
2
)
∫
d
x
(
cos
x
+
sin
x
)
n
−
2
)
{\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left({\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right)}
∫
cos
a
x
d
x
cos
a
x
+
sin
a
x
=
x
2
+
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
{\displaystyle \int {\frac {\cos ax\;dx}{\cos ax+\sin ax}}={\frac {x}{2}}+{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C}
∫
cos
a
x
d
x
cos
a
x
−
sin
a
x
=
x
2
−
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
{\displaystyle \int {\frac {\cos ax\;dx}{\cos ax-\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C}
∫
sin
a
x
d
x
cos
a
x
+
sin
a
x
=
x
2
−
1
2
a
ln
|
sin
a
x
+
cos
a
x
|
+
C
{\displaystyle \int {\frac {\sin ax\;dx}{\cos ax+\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C}
∫
sin
a
x
d
x
cos
a
x
−
sin
a
x
=
−
x
2
−
1
2
a
ln
|
sin
a
x
−
cos
a
x
|
+
C
{\displaystyle \int {\frac {\sin ax\;dx}{\cos ax-\sin ax}}=-{\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C}
∫
cos
a
x
d
x
sin
a
x
(
1
+
cos
a
x
)
=
−
1
4
a
tan
2
a
x
2
+
1
2
a
ln
|
tan
a
x
2
|
+
C
{\displaystyle \int {\frac {\cos ax\;dx}{\sin ax(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}
∫
cos
a
x
d
x
sin
a
x
(
1
+
−
cos
a
x
)
=
−
1
4
a
cot
2
a
x
2
−
1
2
a
ln
|
tan
a
x
2
|
+
C
{\displaystyle \int {\frac {\cos ax\;dx}{\sin ax(1+-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}
∫
sin
a
x
d
x
cos
a
x
(
1
+
sin
a
x
)
=
1
4
a
cot
2
(
a
x
2
+
π
4
)
+
1
2
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
{\displaystyle \int {\frac {\sin ax\;dx}{\cos ax(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}
∫
sin
a
x
d
x
cos
a
x
(
1
−
sin
a
x
)
=
1
4
a
tan
2
(
a
x
2
+
π
4
)
−
1
2
a
ln
|
tan
(
a
x
2
+
π
4
)
|
+
C
{\displaystyle \int {\frac {\sin ax\;dx}{\cos ax(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}
∫
sin
a
x
cos
a
x
d
x
=
1
2
a
sin
2
a
x
+
C
{\displaystyle \int \sin ax\cos ax\;dx={\frac {1}{2a}}\sin ^{2}ax+C\,\!}
∫
sin
a
1
x
cos
a
2
x
d
x
=
−
cos
(
a
1
+
a
2
)
x
2
(
a
1
+
a
2
)
−
cos
(
a
1
−
a
2
)
x
2
(
a
1
−
a
2
)
+
C
(for
|
a
1
|
≠
|
a
2
|
)
{\displaystyle \int \sin a_{1}x\cos a_{2}x\;dx=-{\frac {\cos(a_{1}+a_{2})x}{2(a_{1}+a_{2})}}-{\frac {\cos(a_{1}-a_{2})x}{2(a_{1}-a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!}
∫
sin
n
a
x
cos
a
x
d
x
=
1
a
(
n
+
1
)
sin
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int \sin ^{n}ax\cos ax\;dx={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫
sin
a
x
cos
n
a
x
d
x
=
−
1
a
(
n
+
1
)
cos
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int \sin ax\cos ^{n}ax\;dx=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫
sin
n
a
x
cos
m
a
x
d
x
=
−
sin
n
−
1
a
x
cos
m
+
1
a
x
a
(
n
+
m
)
+
n
−
1
n
+
m
∫
sin
n
−
2
a
x
cos
m
a
x
d
x
(for
m
,
n
>
0
)
{\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;dx=-{\frac {\sin ^{n-1}ax\cos ^{m+1}ax}{a(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}ax\cos ^{m}ax\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}
also:
∫
sin
n
a
x
cos
m
a
x
d
x
=
sin
n
+
1
a
x
cos
m
−
1
a
x
a
(
n
+
m
)
+
m
−
1
n
+
m
∫
sin
n
a
x
cos
m
−
2
a
x
d
x
(for
m
,
n
>
0
)
{\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;dx={\frac {\sin ^{n+1}ax\cos ^{m-1}ax}{a(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}ax\cos ^{m-2}ax\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}
∫
d
x
sin
a
x
cos
a
x
=
1
a
ln
|
tan
a
x
|
+
C
{\displaystyle \int {\frac {dx}{\sin ax\cos ax}}={\frac {1}{a}}\ln \left|\tan ax\right|+C}
∫
d
x
sin
a
x
cos
n
a
x
=
1
a
(
n
−
1
)
cos
n
−
1
a
x
+
∫
d
x
sin
a
x
cos
n
−
2
a
x
(for
n
≠
1
)
{\displaystyle \int {\frac {dx}{\sin ax\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {dx}{\sin ax\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
sin
n
a
x
cos
a
x
=
−
1
a
(
n
−
1
)
sin
n
−
1
a
x
+
∫
d
x
sin
n
−
2
a
x
cos
a
x
(for
n
≠
1
)
{\displaystyle \int {\frac {dx}{\sin ^{n}ax\cos ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {dx}{\sin ^{n-2}ax\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
a
x
d
x
cos
n
a
x
=
1
a
(
n
−
1
)
cos
n
−
1
a
x
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ax\;dx}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
2
a
x
d
x
cos
a
x
=
−
1
a
sin
a
x
+
1
a
ln
|
tan
(
π
4
+
a
x
2
)
|
+
C
{\displaystyle \int {\frac {\sin ^{2}ax\;dx}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {ax}{2}}\right)\right|+C}
∫
sin
2
a
x
d
x
cos
n
a
x
=
sin
a
x
a
(
n
−
1
)
cos
n
−
1
a
x
−
1
n
−
1
∫
d
x
cos
n
−
2
a
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{2}ax\;dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
a
x
d
x
cos
a
x
=
−
sin
n
−
1
a
x
a
(
n
−
1
)
+
∫
sin
n
−
2
a
x
d
x
cos
a
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\;dx}{\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
a
x
d
x
cos
m
a
x
=
sin
n
+
1
a
x
a
(
m
−
1
)
cos
m
−
1
a
x
−
n
−
m
+
2
m
−
1
∫
sin
n
a
x
d
x
cos
m
−
2
a
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}={\frac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}ax\;dx}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
also:
∫
sin
n
a
x
d
x
cos
m
a
x
=
−
sin
n
−
1
a
x
a
(
n
−
m
)
cos
m
−
1
a
x
+
n
−
1
n
−
m
∫
sin
n
−
2
a
x
d
x
cos
m
a
x
(for
m
≠
n
)
{\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}=-{\frac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}ax\;dx}{\cos ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}
also:
∫
sin
n
a
x
d
x
cos
m
a
x
=
sin
n
−
1
a
x
a
(
m
−
1
)
cos
m
−
1
a
x
−
n
−
1
m
−
1
∫
sin
n
−
2
a
x
d
x
cos
m
−
2
a
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}={\frac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}ax\;dx}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
∫
cos
a
x
d
x
sin
n
a
x
=
−
1
a
(
n
−
1
)
sin
n
−
1
a
x
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos ax\;dx}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
cos
2
a
x
d
x
sin
a
x
=
1
a
(
cos
a
x
+
ln
|
tan
a
x
2
|
)
+
C
{\displaystyle \int {\frac {\cos ^{2}ax\;dx}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left|\tan {\frac {ax}{2}}\right|\right)+C}
∫
cos
2
a
x
d
x
sin
n
a
x
=
−
1
n
−
1
(
cos
a
x
a
sin
n
−
1
a
x
)
+
∫
d
x
sin
n
−
2
a
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos ^{2}ax\;dx}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax)}}+\int {\frac {dx}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
cos
n
a
x
d
x
sin
m
a
x
=
−
cos
n
+
1
a
x
a
(
m
−
1
)
sin
m
−
1
a
x
−
n
−
m
−
2
m
−
1
∫
cos
n
a
x
d
x
sin
m
−
2
a
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}=-{\frac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-m-2}{m-1}}\int {\frac {\cos ^{n}ax\;dx}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
also:
∫
cos
n
a
x
d
x
sin
m
a
x
=
cos
n
−
1
a
x
a
(
n
−
m
)
sin
m
−
1
a
x
+
n
−
1
n
−
m
∫
cos
n
−
2
a
x
d
x
sin
m
a
x
(for
m
≠
n
)
{\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}={\frac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\cos ^{n-2}ax\;dx}{\sin ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}
also:
∫
cos
n
a
x
d
x
sin
m
a
x
=
−
cos
n
−
1
a
x
a
(
m
−
1
)
sin
m
−
1
a
x
−
n
−
1
m
−
1
∫
cos
n
−
2
a
x
d
x
sin
m
−
2
a
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}=-{\frac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\cos ^{n-2}ax\;dx}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
∫
sin
a
x
tan
a
x
d
x
=
1
a
(
ln
|
sec
a
x
+
tan
a
x
|
−
sin
a
x
)
+
C
{\displaystyle \int \sin ax\tan ax\;dx={\frac {1}{a}}(\ln |\sec ax+\tan ax|-\sin ax)+C\,\!}
∫
tan
n
a
x
d
x
sin
2
a
x
=
1
a
(
n
−
1
)
tan
n
−
1
(
a
x
)
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\tan ^{n}ax\;dx}{\sin ^{2}ax}}={\frac {1}{a(n-1)}}\tan ^{n-1}(ax)+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
tan
n
a
x
d
x
cos
2
a
x
=
1
a
(
n
+
1
)
tan
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int {\frac {\tan ^{n}ax\;dx}{\cos ^{2}ax}}={\frac {1}{a(n+1)}}\tan ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫
cot
n
a
x
d
x
sin
2
a
x
=
1
a
(
n
+
1
)
cot
n
+
1
a
x
+
C
(for
n
≠
−
1
)
{\displaystyle \int {\frac {\cot ^{n}ax\;dx}{\sin ^{2}ax}}={\frac {1}{a(n+1)}}\cot ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫
cot
n
a
x
d
x
cos
2
a
x
=
1
a
(
1
−
n
)
tan
1
−
n
a
x
+
C
(for
n
≠
1
)
{\displaystyle \int {\frac {\cot ^{n}ax\;dx}{\cos ^{2}ax}}={\frac {1}{a(1-n)}}\tan ^{1-n}ax+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
−
c
c
sin
x
d
x
=
0
{\displaystyle \int _{-c}^{c}\sin {x}\;dx=0\!}
∫
−
c
c
cos
x
d
x
=
2
∫
0
c
cos
x
d
x
=
2
∫
−
c
0
cos
x
d
x
=
2
sin
c
{\displaystyle \int _{-c}^{c}\cos {x}\;dx=2\int _{0}^{c}\cos {x}\;dx=2\int _{-c}^{0}\cos {x}\;dx=2\sin {c}\!}
∫
−
c
c
tan
x
d
x
=
0
{\displaystyle \int _{-c}^{c}\tan {x}\;dx=0\!}
∫
−
a
2
a
2
x
2
cos
2
n
π
x
a
d
x
=
a
3
(
n
2
π
2
−
6
)
24
n
2
π
2
(for
n
=
1
,
3
,
5...
)
{\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=1,3,5...{\mbox{)}}\,\!}
∫
arcsin
x
c
d
x
=
x
arcsin
x
c
+
c
2
−
x
2
{\displaystyle \int \arcsin {\frac {x}{c}}\,dx=x\arcsin {\frac {x}{c}}+{\sqrt {c^{2}-x^{2}}}}
∫
x
arcsin
x
c
d
x
=
(
x
2
2
−
c
2
4
)
arcsin
x
c
+
x
4
c
2
−
x
2
{\displaystyle \int x\arcsin {\frac {x}{c}}\,dx=\left({\frac {x^{2}}{2}}-{\frac {c^{2}}{4}}\right)\arcsin {\frac {x}{c}}+{\frac {x}{4}}{\sqrt {c^{2}-x^{2}}}}
∫
x
2
arcsin
x
c
d
x
=
x
3
3
arcsin
x
c
+
x
2
+
2
c
2
9
c
2
−
x
2
{\displaystyle \int x^{2}\arcsin {\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\arcsin {\frac {x}{c}}+{\frac {x^{2}+2c^{2}}{9}}{\sqrt {c^{2}-x^{2}}}}
∫
x
n
sin
−
1
x
d
x
=
1
n
+
1
(
x
n
+
1
sin
−
1
x
{\displaystyle \int x^{n}\sin ^{-1}x\,dx={\frac {1}{n+1}}\left(x^{n+1}\sin ^{-1}x\right.}
+
x
n
1
−
x
2
−
n
x
n
−
1
sin
−
1
x
n
−
1
+
n
∫
x
n
−
2
sin
−
1
x
d
x
)
{\displaystyle \left.+{\frac {x^{n}{\sqrt {1-x^{2}}}-nx^{n-1}\sin ^{-1}x}{n-1}}+n\int x^{n-2}\sin ^{-1}x\,dx\right)}
∫
arccos
x
c
d
x
=
x
arccos
x
c
−
c
2
−
x
2
{\displaystyle \int \arccos {\frac {x}{c}}\,dx=x\arccos {\frac {x}{c}}-{\sqrt {c^{2}-x^{2}}}}
∫
x
arccos
x
c
d
x
=
(
x
2
2
−
c
2
4
)
arccos
x
c
−
x
4
c
2
−
x
2
{\displaystyle \int x\arccos {\frac {x}{c}}\,dx=\left({\frac {x^{2}}{2}}-{\frac {c^{2}}{4}}\right)\arccos {\frac {x}{c}}-{\frac {x}{4}}{\sqrt {c^{2}-x^{2}}}}
∫
x
2
arccos
x
c
d
x
=
x
3
3
arccos
x
c
−
x
2
+
2
c
2
9
c
2
−
x
2
{\displaystyle \int x^{2}\arccos {\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\arccos {\frac {x}{c}}-{\frac {x^{2}+2c^{2}}{9}}{\sqrt {c^{2}-x^{2}}}}
∫
arctan
x
c
d
x
=
x
arctan
x
c
−
c
2
ln
(
c
2
+
x
2
)
{\displaystyle \int \arctan {\frac {x}{c}}\,dx=x\arctan {\frac {x}{c}}-{\frac {c}{2}}\ln(c^{2}+x^{2})}
∫
x
arctan
x
c
d
x
=
c
2
+
x
2
2
arctan
x
c
−
c
x
2
{\displaystyle \int x\arctan {\frac {x}{c}}\,dx={\frac {c^{2}+x^{2}}{2}}\arctan {\frac {x}{c}}-{\frac {cx}{2}}}
∫
x
2
arctan
x
c
d
x
=
x
3
3
arctan
x
c
−
c
x
2
6
+
c
3
6
ln
c
2
+
x
2
{\displaystyle \int x^{2}\arctan {\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\arctan {\frac {x}{c}}-{\frac {cx^{2}}{6}}+{\frac {c^{3}}{6}}\ln {c^{2}+x^{2}}}
∫
x
n
arctan
x
c
d
x
=
x
n
+
1
n
+
1
arctan
x
c
−
c
n
+
1
∫
x
n
+
1
d
x
c
2
+
x
2
(
n
≠
1
)
{\displaystyle \int x^{n}\arctan {\frac {x}{c}}\,dx={\frac {x^{n+1}}{n+1}}\arctan {\frac {x}{c}}-{\frac {c}{n+1}}\int {\frac {x^{n+1}dx}{c^{2}+x^{2}}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}
∫
arcsec
x
c
d
x
=
x
arcsec
x
c
+
x
c
|
x
|
ln
|
x
±
x
2
−
1
|
{\displaystyle \int \operatorname {arcsec} {\frac {x}{c}}\,dx=x\operatorname {arcsec} {\frac {x}{c}}+{\frac {x}{c|x|}}\ln {|x\pm {\sqrt {x^{2}-1}}|}}
∫
x
arcsec
x
d
x
=
1
2
(
x
2
arcsec
x
−
x
2
−
1
)
{\displaystyle \int x\operatorname {arcsec} {x}\,dx\,=\,{\frac {1}{2}}\left(x^{2}\operatorname {arcsec} {x}-{\sqrt {x^{2}-1}}\right)}
∫
x
n
arcsec
x
d
x
=
1
n
+
1
(
x
n
+
1
arcsec
x
−
1
n
(
x
n
−
1
x
2
−
1
{\displaystyle \int x^{n}\operatorname {arcsec} {x}\,dx\,=\,{\frac {1}{n+1}}\left(x^{n+1}\operatorname {arcsec} {x}-{\frac {1}{n}}\left(x^{n-1}{\sqrt {x^{2}-1}}\;\right.\right.}
+
(
1
−
n
)
(
x
n
−
1
arcsec
x
+
(
1
−
n
)
∫
x
n
−
2
arcsec
x
d
x
)
)
)
{\displaystyle \left.\left.+(1-n)\left(x^{n-1}\operatorname {arcsec} {x}+(1-n)\int x^{n-2}\operatorname {arcsec} {x}\,dx\right)\right)\right)}
∫
a
r
c
c
o
t
x
c
d
x
=
x
a
r
c
c
o
t
x
c
+
c
2
ln
(
c
2
+
x
2
)
{\displaystyle \int \mathrm {arccot} \,{\frac {x}{c}}\,dx=x\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {c}{2}}\ln(c^{2}+x^{2})}
∫
x
a
r
c
c
o
t
x
c
d
x
=
c
2
+
x
2
2
a
r
c
c
o
t
x
c
+
c
x
2
{\displaystyle \int x\,\mathrm {arccot} \,{\frac {x}{c}}\,dx={\frac {c^{2}+x^{2}}{2}}\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {cx}{2}}}
∫
x
2
a
r
c
c
o
t
x
c
d
x
=
x
3
3
a
r
c
c
o
t
x
c
+
c
x
2
6
−
c
3
6
ln
(
c
2
+
x
2
)
{\displaystyle \int x^{2}\,\mathrm {arccot} \,{\frac {x}{c}}\,dx={\frac {x^{3}}{3}}\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {cx^{2}}{6}}-{\frac {c^{3}}{6}}\ln(c^{2}+x^{2})}
∫
x
n
a
r
c
c
o
t
x
c
d
x
=
x
n
+
1
n
+
1
a
r
c
c
o
t
x
c
+
c
n
+
1
∫
x
n
+
1
d
x
c
2
+
x
2
(
n
≠
1
)
{\displaystyle \int x^{n}\,\mathrm {arccot} \,{\frac {x}{c}}\,dx={\frac {x^{n+1}}{n+1}}\,\mathrm {arccot} \,{\frac {x}{c}}+{\frac {c}{n+1}}\int {\frac {x^{n+1}dx}{c^{2}+x^{2}}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
(
n
≠
−
1
)
{\displaystyle \int (ax+b)^{n}dx={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{( }}n\neq -1{\mbox{)}}\,\!}
∫
d
x
a
x
+
b
=
1
a
ln
|
a
x
+
b
|
{\displaystyle \int {\frac {dx}{ax+b}}={\frac {1}{a}}\ln \left|ax+b\right|}
∫
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
}
)
{\displaystyle \int x(ax+b)^{n}dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{( }}n\not \in \{1,2\}{\mbox{)}}}
∫
x
a
x
+
b
d
x
=
x
a
−
b
a
2
ln
|
a
x
+
b
|
{\displaystyle \int {\frac {x}{ax+b}}dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|}
∫
x
(
a
x
+
b
)
2
d
x
=
b
a
2
(
a
x
+
b
)
+
1
a
2
ln
|
a
x
+
b
|
{\displaystyle \int {\frac {x}{(ax+b)^{2}}}dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|}
∫
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}}}\qquad {\mbox{( }}n\not \in \{1,2\}{\mbox{)}}}
∫
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
+
b
)
n
−
2
−
b
2
(
n
−
1
)
(
a
x
+
b
)
n
−
1
)
(
n
∉
{
1
,
2
,
3
}
)
{\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)(a+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right)\qquad {\mbox{( }}n\not \in \{1,2,3\}{\mbox{)}}}
∫
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
x
2
+
a
2
=
1
a
arctan
x
a
{\displaystyle \int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!}
∫
d
x
x
2
−
a
2
=
−
1
a
a
r
c
t
a
n
h
x
a
=
1
2
a
ln
a
−
x
a
+
x
(
|
x
|
<
|
a
|
)
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}\qquad {\mbox{( }}|x|<|a|{\mbox{)}}\,\!}
∫
d
x
x
2
−
a
2
=
−
1
a
a
r
c
c
o
t
h
x
a
=
1
2
a
ln
x
−
a
x
+
a
(
|
x
|
>
|
a
|
)
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}\qquad {\mbox{( }}|x|>|a|{\mbox{)}}\,\!}
∫
d
x
a
x
2
+
b
x
+
c
=
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
(
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{( }}4ac-b^{2}>0{\mbox{)}}}
∫
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
|
(
4
a
c
−
b
2
<
0
)
{\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|\qquad {\mbox{( }}4ac-b^{2}<0{\mbox{)}}}
∫
d
x
a
x
2
+
b
x
+
c
=
−
2
2
a
x
+
b
(
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}=-{\frac {2}{2ax+b}}\qquad {\mbox{( }}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
(
4
a
c
−
b
2
>
0
)
{\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}}}}\qquad {\mbox{( }}4ac-b^{2}>0{\mbox{)}}}
∫
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
(
4
a
c
−
b
2
<
0
)
{\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}}}\qquad {\mbox{( }}4ac-b^{2}<0{\mbox{)}}}
∫
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
)
(
4
a
c
−
b
2
=
0
)
{\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)}}\qquad {\mbox{( }}4ac-b^{2}=0{\mbox{)}}}
∫
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}}}
∫
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
+
3
8
a
2
x
r
+
3
8
a
4
ln
(
x
+
r
)
{\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {3}{8}}a^{2}xr+{\frac {3}{8}}a^{4}\ln \left(x+r\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
)
{\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(x+r\right)}
∫
x
r
d
x
=
r
3
3
{\displaystyle \int xr\;dx={\frac {r^{3}}{3}}}
∫
x
r
3
d
x
=
r
5
5
{\displaystyle \int xr^{3}\;dx={\frac {r^{5}}{5}}}
∫
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
)
{\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left(x+r\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
)
{\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(x+r\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
)
{\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(x+r\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
)
{\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(x+r\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
arsinh
a
x
{\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\,\operatorname {arsinh} {\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
=
arsinh
x
a
=
ln
(
x
+
r
a
)
{\displaystyle \int {\frac {dx}{r}}=\operatorname {arsinh} {\frac {x}{a}}=\ln \left({\frac {x+r}{a}}\right)}
∫
d
x
r
3
=
x
a
2
r
{\displaystyle \int {\frac {dx}{r^{3}}}={\frac {x}{a^{2}r}}}
∫
x
d
x
r
=
r
{\displaystyle \int {\frac {x\,dx}{r}}=r}
∫
x
d
x
r
3
=
−
1
r
{\displaystyle \int {\frac {x\,dx}{r^{3}}}=-{\frac {1}{r}}}
∫
x
2
d
x
r
=
x
2
r
−
a
2
2
arsinh
x
a
=
x
2
r
−
a
2
2
ln
(
x
+
r
a
)
{\displaystyle \int {\frac {x^{2}\;dx}{r}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\,\operatorname {arsinh} {\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left({\frac {x+r}{a}}\right)}
∫
d
x
x
r
=
−
1
a
arsinh
a
x
=
−
1
a
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\operatorname {arsinh} {\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|}
Category
:
Tích phân
Hidden category:
VI