Dit artikel bevat een lijst van integralen van irrationale functies . Integralen zijn het onderwerp van studie van de integraalrekening . De integralen in de lijst hieronder zijn veel voorkomende integralen van een functie onder de wortel . Er wordt van alle integralen de primitieve functie gegeven, maar de integratieconstante is in de uitkomst steeds weggelaten.
r
=
x
2
+
a
2
{\displaystyle r={\sqrt {x^{2}+a^{2}}}}
bewerken
∫
r
d
x
=
1
2
(
x
r
+
a
2
ln
(
x
+
r
)
)
{\displaystyle \int r\ \mathrm {d} x={\tfrac {1}{2}}(xr+a^{2}\ \ln(x+r))}
∫
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}\ \mathrm {d} x={\tfrac {1}{4}}xr^{3}+{\tfrac {3}{8}}a^{2}xr+{\tfrac {3}{8}}a^{4}\ln(x+r)}
∫
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}\ \mathrm {d} x={\tfrac {1}{6}}xr^{5}+{\tfrac {5}{24}}a^{2}xr^{3}+{\tfrac {5}{16}}a^{4}xr+{\tfrac {5}{16}}a^{6}\ln(x+r)}
∫
x
r
d
x
=
1
3
r
3
{\displaystyle \int xr\ \mathrm {d} x={\tfrac {1}{3}}r^{3}}
∫
x
r
3
d
x
=
1
5
r
5
{\displaystyle \int xr^{3}\ \mathrm {d} x={\tfrac {1}{5}}r^{5}}
∫
x
r
2
n
+
1
d
x
=
r
2
n
+
3
2
n
+
3
{\displaystyle \int xr^{2n+1}\ \mathrm {d} x={\frac {r^{2n+3}}{2n+3}}}
∫
x
2
r
d
x
=
1
4
x
r
3
−
1
8
a
2
x
r
−
1
8
a
4
ln
(
x
+
r
)
{\displaystyle \int x^{2}r\ \mathrm {d} x={\tfrac {1}{4}}xr^{3}-{\tfrac {1}{8}}a^{2}xr-{\tfrac {1}{8}}a^{4}\ln(x+r)}
∫
x
2
r
3
d
x
=
1
6
x
r
5
−
1
24
a
2
x
r
3
−
1
16
a
4
x
r
−
1
16
a
6
ln
(
x
+
r
)
{\displaystyle \int x^{2}r^{3}\ \mathrm {d} x={\tfrac {1}{6}}xr^{5}-{\tfrac {1}{24}}a^{2}xr^{3}-{\tfrac {1}{16}}a^{4}xr-{\tfrac {1}{16}}a^{6}\ln(x+r)}
∫
x
3
r
d
x
=
1
5
r
5
−
1
3
a
2
r
3
{\displaystyle \int x^{3}r\ \mathrm {d} x={\tfrac {1}{5}}r^{5}-{\tfrac {1}{3}}a^{2}r^{3}}
∫
x
3
r
3
d
x
=
1
7
r
7
−
1
5
a
2
r
5
{\displaystyle \int x^{3}r^{3}\ \mathrm {d} x={\tfrac {1}{7}}r^{7}-{\tfrac {1}{5}}a^{2}r^{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}\ \mathrm {d} x={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}}
∫
x
4
r
d
x
=
1
6
x
3
r
3
−
1
8
a
2
x
r
3
+
1
16
a
4
x
r
+
1
16
a
6
ln
(
x
+
r
)
{\displaystyle \int x^{4}r\ \mathrm {d} x={\tfrac {1}{6}}x^{3}r^{3}-{\tfrac {1}{8}}a^{2}xr^{3}+{\tfrac {1}{16}}a^{4}xr+{\tfrac {1}{16}}a^{6}\ln(x+r)}
∫
x
4
r
3
d
x
=
1
8
x
3
r
5
−
1
16
a
2
x
r
5
+
1
64
a
4
x
r
3
+
3
128
a
6
x
r
+
3
128
a
8
ln
(
x
+
r
)
{\displaystyle \int x^{4}r^{3}\ \mathrm {d} x={\tfrac {1}{8}}x^{3}r^{5}-{\tfrac {1}{16}}a^{2}xr^{5}+{\tfrac {1}{64}}a^{4}xr^{3}+{\tfrac {3}{128}}a^{6}xr+{\tfrac {3}{128}}a^{8}\ln(x+r)}
∫
x
5
r
d
x
=
1
7
r
7
−
2
5
a
2
r
5
+
1
3
a
4
r
3
{\displaystyle \int x^{5}r\ \mathrm {d} x={\tfrac {1}{7}}r^{7}-{\tfrac {2}{5}}a^{2}r^{5}+{\tfrac {1}{3}}a^{4}r^{3}}
∫
x
5
r
3
d
x
=
1
9
r
9
−
2
7
a
2
r
7
+
1
5
a
4
r
5
{\displaystyle \int x^{5}r^{3}\ \mathrm {d} x={\tfrac {1}{9}}r^{9}-{\tfrac {2}{7}}a^{2}r^{7}+{\tfrac {1}{5}}a^{4}r^{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}\ \mathrm {d} x={\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\ \mathrm {d} x}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\ \operatorname {arsinh} {\frac {a}{x}}}
∫
r
3
d
x
x
=
1
3
r
3
+
a
2
r
−
a
3
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {r^{3}\ \mathrm {d} x}{x}}={\tfrac {1}{3}}r^{3}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|}
∫
r
5
d
x
x
=
1
5
r
5
+
1
3
a
2
r
3
+
a
4
r
−
a
5
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {r^{5}\ \mathrm {d} x}{x}}={\tfrac {1}{5}}r^{5}+{\tfrac {1}{3}}a^{2}r^{3}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}}\right|}
∫
r
7
d
x
x
=
1
7
r
7
+
1
5
a
2
r
5
+
1
3
a
4
r
3
+
a
6
r
−
a
7
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {r^{7}\ \mathrm {d} x}{x}}={\tfrac {1}{7}}r^{7}+{\tfrac {1}{5}}a^{2}r^{5}+{\tfrac {1}{3}}a^{4}r^{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 {\mathrm {d} x}{r}}=\operatorname {arsinh} {\frac {x}{a}}=\ln \left({\frac {x+r}{a}}\right)}
∫
d
x
r
3
=
x
a
2
r
{\displaystyle \int {\frac {\mathrm {d} x}{r^{3}}}={\frac {x}{a^{2}r}}}
∫
x
d
x
r
=
r
{\displaystyle \int {\frac {x\ \ \mathrm {d} x}{r}}=r}
∫
x
d
x
r
3
=
−
1
r
{\displaystyle \int {\frac {x\ \ \mathrm {d} x}{r^{3}}}=-{\frac {1}{r}}}
∫
x
2
d
x
r
=
1
2
x
r
−
1
2
a
2
arsinh
x
a
=
1
2
x
r
−
1
2
a
2
ln
(
x
+
r
a
)
{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{r}}={\tfrac {1}{2}}xr-{\tfrac {1}{2}}a^{2}\ \operatorname {arsinh} {\frac {x}{a}}={\tfrac {1}{2}}xr-{\tfrac {1}{2}}a^{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 {\mathrm {d} x}{xr}}=-{\frac {1}{a}}\ \operatorname {arsinh} {\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|}
s
=
x
2
−
a
2
{\displaystyle s={\sqrt {x^{2}-a^{2}}}}
bewerken
Voor de volgende integralen is
x
2
>
a
2
{\displaystyle x^{2}>a^{2}}
∫
s
d
x
=
1
2
(
x
s
−
a
2
ln
(
x
+
s
)
)
{\displaystyle \int s\ \mathrm {d} x={\tfrac {1}{2}}(xs-a^{2}\ln(x+s))}
∫
x
s
d
x
=
1
3
s
3
{\displaystyle \int xs\ \mathrm {d} x={\tfrac {1}{3}}s^{3}}
∫
s
d
x
x
=
s
−
a
arccos
|
a
x
|
{\displaystyle \int {\frac {s\ \mathrm {d} x}{x}}=s-a\arccos \left|{\frac {a}{x}}\right|}
∫
d
x
s
=
∫
d
x
x
2
−
a
2
=
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {\mathrm {d} x}{s}}=\int {\frac {\mathrm {d} x}{\sqrt {x^{2}-a^{2}}}}=\ln \left|{\frac {x+s}{a}}\right|}
ln
|
x
+
s
a
|
=
s
g
n
(
x
)
arcosh
|
x
a
|
=
1
2
ln
(
x
+
s
x
−
s
)
{\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\mathrm {sgn} (x)\ \operatorname {arcosh} \left|{\frac {x}{a}}\right|={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)}
arcosh
|
x
a
|
{\displaystyle \operatorname {arcosh} \left|{\frac {x}{a}}\right|}
∫
x
d
x
s
=
s
{\displaystyle \int {\frac {x\ \mathrm {d} x}{s}}=s}
∫
x
d
x
s
3
=
−
1
s
{\displaystyle \int {\frac {x\ \mathrm {d} x}{s^{3}}}=-{\frac {1}{s}}}
∫
x
d
x
s
5
=
−
1
3
s
3
{\displaystyle \int {\frac {x\ \mathrm {d} x}{s^{5}}}=-{\frac {1}{3s^{3}}}}
∫
x
d
x
s
7
=
−
1
5
s
5
{\displaystyle \int {\frac {x\ \mathrm {d} x}{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\ \mathrm {d} x}{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}\ \mathrm {d} x}{s^{2n+1}}}=-{\frac {1}{2n-1}}{\frac {x^{2m-1}}{s^{2n-1}}}+{\frac {2m-1}{2n-1}}\int {\frac {x^{2m-2}\ \mathrm {d} x}{s^{2n-1}}}}
∫
x
2
d
x
s
=
1
2
x
s
+
1
2
a
2
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s}}={\tfrac {1}{2}}xs+{\tfrac {1}{2}}a^{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}\ \mathrm {d} x}{s^{3}}}=-{\frac {x}{s}}+\ln \left|{\frac {x+s}{a}}\right|}
∫
x
4
d
x
s
=
1
4
x
3
s
+
3
8
a
2
x
s
+
3
8
a
4
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{4}\ \mathrm {d} x}{s}}={\tfrac {1}{4}}x^{3}s+{\tfrac {3}{8}}a^{2}xs+{\tfrac {3}{8}}a^{4}\ln \left|{\frac {x+s}{a}}\right|}
∫
x
4
d
x
s
3
=
1
2
x
s
−
a
2
x
s
+
3
2
a
2
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{4}\ \mathrm {d} x}{s^{3}}}={\tfrac {1}{2}}xs-{\frac {a^{2}x}{s}}+{\tfrac {3}{2}}a^{2}\ln \left|{\frac {x+s}{a}}\right|}
∫
x
4
d
x
s
5
=
−
x
s
−
x
3
3
s
3
+
ln
|
x
+
s
a
|
{\displaystyle \int {\frac {x^{4}\ \mathrm {d} x}{s^{5}}}=-{\frac {x}{s}}-{\frac {x^{3}}{3s^{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
n
>
m
≥
0
{\displaystyle \int {\frac {x^{2m}\ \mathrm {d} x}{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}}}\qquad n>m\geq 0}
∫
d
x
s
3
=
−
1
a
2
x
s
{\displaystyle \int {\frac {\mathrm {d} x}{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 {\mathrm {d} x}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\tfrac {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 {\mathrm {d} x}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\tfrac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {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 {\mathrm {d} x}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\tfrac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\tfrac {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}\ \mathrm {d} x}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}}
∫
x
2
d
x
s
7
=
1
a
4
[
1
3
x
3
3
s
3
−
1
5
x
5
s
5
]
{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{7}}}={\frac {1}{a^{4}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{3s^{3}}}-{\tfrac {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}\ \mathrm {d} x}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\tfrac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}
u
=
a
2
−
x
2
{\displaystyle u={\sqrt {a^{2}-x^{2}}}}
bewerken
∫
u
d
x
=
1
2
(
x
u
+
a
2
arcsin
x
a
)
|
x
|
≤
|
a
|
{\displaystyle \int u\ \mathrm {d} x={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad |x|\leq |a|}
∫
x
u
d
x
=
−
1
3
u
3
(
|
x
|
≤
|
a
|
)
{\displaystyle \int xu\ \mathrm {d} x=-{\frac {1}{3}}u^{3}\qquad (|x|\leq |a|)}
∫
x
2
u
d
x
=
−
x
4
u
3
+
a
2
8
(
x
u
+
a
2
arcsin
x
a
)
|
x
|
≤
|
a
|
{\displaystyle \int x^{2}u\ \mathrm {d} x=-{\frac {x}{4}}u^{3}+{\frac {a^{2}}{8}}(xu+a^{2}\arcsin {\frac {x}{a}})\qquad |x|\leq |a|}
∫
u
d
x
x
=
u
−
a
ln
|
a
+
u
x
|
|
x
|
≤
|
a
|
{\displaystyle \int {\frac {u\ \mathrm {d} x}{x}}=u-a\ln \left|{\frac {a+u}{x}}\right|\qquad |x|\leq |a|}
∫
d
x
u
=
arcsin
x
a
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {\mathrm {d} x}{u}}=\arcsin {\frac {x}{a}}\qquad (|x|\leq |a|)}
∫
x
2
d
x
u
=
1
2
(
−
x
u
+
a
2
arcsin
x
a
)
|
x
|
≤
|
a
|
{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{u}}={\frac {1}{2}}\left(-xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad |x|\leq |a|}
∫
u
d
x
=
1
2
(
x
u
−
sgn
x
arcosh
|
x
a
|
)
voor
|
x
|
≥
|
a
|
{\displaystyle \int u\ \mathrm {d} x={\frac {1}{2}}\left(xu-\operatorname {sgn} x\ \operatorname {arcosh} \left|{\frac {x}{a}}\right|\right)\qquad {\mbox{voor }}|x|\geq |a|}
∫
x
u
d
x
=
−
u
|
x
|
≤
|
a
|
{\displaystyle \int {\frac {x}{u}}\ \mathrm {d} x=-u\qquad |x|\leq |a|}
R
=
a
x
2
+
b
x
+
c
{\displaystyle R={\sqrt {ax^{2}+bx+c}}}
bewerken
Neem aan dat
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
(
p
x
+
q
)
2
{\displaystyle (px+q)^{2}}
∫
d
x
R
=
1
a
ln
|
2
a
R
+
2
a
x
+
b
|
voor
a
>
0
{\displaystyle \int {\frac {\mathrm {d} x}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|\qquad {\mbox{voor }}a>0}
∫
d
x
R
=
1
a
arsinh
2
a
x
+
b
4
a
c
−
b
2
voor
a
>
0
,
4
a
c
−
b
2
>
0
{\displaystyle \int {\frac {\mathrm {d} x}{R}}={\frac {1}{\sqrt {a}}}\ \operatorname {arsinh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{voor }}a>0,\ 4ac-b^{2}>0}
∫
d
x
R
=
1
a
ln
|
2
a
x
+
b
|
voor
a
>
0
,
4
a
c
−
b
2
=
0
{\displaystyle \int {\frac {\mathrm {d} x}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{voor }}a>0,\ 4ac-b^{2}=0}
∫
d
x
R
=
−
1
−
a
arcsin
2
a
x
+
b
b
2
−
4
a
c
voor
a
<
0
,
4
a
c
−
b
2
<
0
,
|
2
a
x
+
b
|
<
b
2
−
4
a
c
{\displaystyle \int {\frac {\mathrm {d} x}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{voor }}a<0,\ 4ac-b^{2}<0,\ |2ax+b|<{\sqrt {b^{2}-4ac}}}
∫
d
x
R
3
=
4
a
x
+
2
b
(
4
a
c
−
b
2
)
R
{\displaystyle \int {\frac {\mathrm {d} x}{R^{3}}}={\frac {4ax+2b}{(4ac-b^{2})R}}}
∫
d
x
R
5
=
4
a
x
+
2
b
3
(
4
a
c
−
b
2
)
R
(
1
R
2
+
8
a
4
a
c
−
b
2
)
{\displaystyle \int {\frac {\mathrm {d} x}{R^{5}}}={\frac {4ax+2b}{3(4ac-b^{2})R}}\left({\frac {1}{R^{2}}}+{\frac {8a}{4ac-b^{2}}}\right)}
∫
d
x
R
2
n
+
1
=
2
(
2
n
−
1
)
(
4
a
c
−
b
2
)
(
2
a
x
+
b
R
2
n
−
1
+
4
a
(
n
−
1
)
∫
d
x
R
2
n
−
1
)
{\displaystyle \int {\frac {\mathrm {d} x}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left({\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {\mathrm {d} x}{R^{2n-1}}}\right)}
∫
x
R
d
x
=
R
a
−
b
2
a
∫
d
x
R
{\displaystyle \int {\frac {x}{R}}\ \mathrm {d} x={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {\mathrm {d} x}{R}}}
∫
x
R
3
d
x
=
−
2
b
x
+
4
c
(
4
a
c
−
b
2
)
R
{\displaystyle \int {\frac {x}{R^{3}}}\ \mathrm {d} x=-{\frac {2bx+4c}{(4ac-b^{2})R}}}
∫
x
R
2
n
+
1
d
x
=
−
1
(
2
n
−
1
)
a
R
2
n
−
1
−
b
2
a
∫
d
x
R
2
n
+
1
{\displaystyle \int {\frac {x}{R^{2n+1}}}\ \mathrm {d} x=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {\mathrm {d} x}{R^{2n+1}}}}
∫
d
x
x
R
=
−
1
c
ln
(
2
c
R
+
b
x
+
2
c
x
)
{\displaystyle \int {\frac {\mathrm {d} x}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right)}
∫
d
x
x
R
=
−
1
c
arsinh
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
{\displaystyle \int {\frac {\mathrm {d} x}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}
S
=
a
x
+
b
{\displaystyle S={\sqrt {ax+b}}}
bewerken
∫
S
d
x
=
2
S
3
3
a
{\displaystyle \int S{\mathrm {d} x}={\frac {2S^{3}}{3a}}}
∫
d
x
S
=
2
S
a
{\displaystyle \int {\frac {\mathrm {d} x}{S}}={\frac {2S}{a}}}
∫
d
x
x
S
=
{
voor
b
>
0
,
a
x
>
0
−
2
b
a
r
t
a
n
h
(
S
b
)
voor
b
>
0
,
a
x
<
0
2
−
b
arctan
(
S
−
b
)
voor
b
<
0
{\displaystyle \int {\frac {\mathrm {d} x}{xS}}={\begin{cases}{\mbox{voor }}b>0,\quad ax>0\\-{\frac {2}{\sqrt {b}}}\mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)&{\mbox{voor }}b>0,\quad ax<0\\{\frac {2}{\sqrt {-b}}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)&{\mbox{voor }}b<0\\\end{cases}}}
∫
S
x
d
x
=
{
2
(
S
−
b
a
r
c
o
t
h
(
S
b
)
)
voor
b
>
0
,
a
x
>
0
2
(
S
−
b
a
r
t
a
n
h
(
S
b
)
)
voor
b
>
0
,
a
x
<
0
2
(
S
−
−
b
arctan
(
S
−
b
)
)
voor
b
<
0
{\displaystyle \int {\frac {S}{x}}\ \mathrm {d} x={\begin{cases}2\left(S-{\sqrt {b}}\ \mathrm {arcoth} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{voor }}b>0,\quad ax>0\\2\left(S-{\sqrt {b}}\ \mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{voor }}b>0,\quad ax<0\\2\left(S-{\sqrt {-b}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)\right)&{\mbox{voor }}b<0\\\end{cases}}}
∫
x
n
S
d
x
=
2
a
(
2
n
+
1
)
(
x
n
S
−
b
n
∫
x
n
−
1
S
d
x
)
{\displaystyle \int {\frac {x^{n}}{S}}\ \mathrm {d} x={\frac {2}{a(2n+1)}}\left(x^{n}S-bn\int {\frac {x^{n-1}}{S}}\ \mathrm {d} x\right)}
∫
x
n
S
d
x
=
2
a
(
2
n
+
3
)
(
x
n
S
3
−
n
b
∫
x
n
−
1
S
d
x
)
{\displaystyle \int x^{n}S\ \mathrm {d} x={\frac {2}{a(2n+3)}}\left(x^{n}S^{3}-nb\int x^{n-1}S\ \mathrm {d} x\right)}
∫
1
x
n
S
d
x
=
−
1
b
(
n
−
1
)
(
S
x
n
−
1
+
(
n
−
3
2
)
a
∫
d
x
x
n
−
1
S
)
{\displaystyle \int {\frac {1}{x^{n}S}}\ \mathrm {d} x=-{\frac {1}{b(n-1)}}\left({\frac {S}{x^{n-1}}}+\left(n-{\frac {3}{2}}\right)a\int {\frac {\mathrm {d} x}{x^{n-1}S}}\right)}
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd705560c62f024beb029c160f94b55c41dc9422)
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\tfrac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a6f3edccad00497079e192ae699e960e152acb4)
![{\displaystyle \int {\frac {\mathrm {d} x}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\tfrac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\tfrac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e712fcf25118a8188eab07ee1cdbcf6fa4ae4659)
![{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{7}}}={\frac {1}{a^{4}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{3s^{3}}}-{\tfrac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8144c9e828d29a8e0537fd6d032a515d6ad8a19a)
![{\displaystyle \int {\frac {x^{2}\ \mathrm {d} x}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\tfrac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\tfrac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\tfrac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bd7215c6045dc7365d33c5c6a50e72d33b8647b)
