7.9
7.3
8.068
7.583
5.966
4
7.55
6.168
7.247
6.803
7.654
6.913
6.6
7.961
7.401
5.8
5.939
7.378
8.565
8.69
6.502
7.478
6.688
6.7
7.2
6.75
8.392
7.719
7.554
5.939
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- Source: Daftar integral dari fungsi irasional
Artikel ini merupakan daftar integral dari fungsi irrasional. Untuk daftar integral lainnya, lihat tabel integral.
r
=
x
2
+
a
2
{\displaystyle r={\sqrt {x^{2}+a^{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
)
{\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {1}{8}}3a^{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
sinh
−
1
a
x
{\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\sinh ^{-1}{\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
=
sinh
−
1
x
a
=
ln
|
x
+
r
|
{\displaystyle \int {\frac {dx}{r}}=\sinh ^{-1}{\frac {x}{a}}=\ln \left|x+r\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
sinh
−
1
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}}\,\sinh ^{-1}{\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left|x+r\right|}
∫
d
x
x
r
=
−
1
a
sinh
−
1
a
x
=
−
1
a
ln
|
a
+
r
x
|
{\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\sinh ^{-1}{\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}}}}
Anggap
(
x
2
>
a
2
)
{\displaystyle (x^{2}>a^{2})}
, untuk
(
x
2
<
a
2
)
{\displaystyle (x^{2}
, perhatikan bagian berikutnya:
∫
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
a
|
{\displaystyle \int {\frac {dx}{s}}=\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\ln \left|{\frac {x+s}{a}}\right|}
Perhatikan bahwa
ln
|
x
+
s
a
|
=
s
g
n
(
x
)
cosh
−
1
|
x
a
|
=
1
2
ln
(
x
+
s
x
−
s
)
{\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\mathrm {sgn} (x)\cosh ^{-1}\left|{\frac {x}{a}}\right|={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)}
, dimana nilai positif dari
cosh
−
1
|
x
a
|
{\displaystyle \cosh ^{-1}\left|{\frac {x}{a}}\right|}
lah yang diambil.
∫
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
(
n
>
m
≥
0
)
{\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}}}\qquad {\mbox{(}}n>m\geq 0{\mbox{)}}}
∫
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]}
t
=
a
2
−
x
2
{\displaystyle t={\sqrt {a^{2}-x^{2}}}}
∫
t
d
x
=
1
2
(
x
t
+
a
2
arcsin
x
a
)
(
|
x
|
≤
|
a
|
)
{\displaystyle \int t\;dx={\frac {1}{2}}\left(xt+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x
t
d
x
=
−
1
3
t
3
(
|
x
|
≤
|
a
|
)
{\displaystyle \int xt\;dx=-{\frac {1}{3}}t^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
t
d
x
x
=
t
−
a
ln
|
a
+
t
x
|
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {t\;dx}{x}}=t-a\ln \left|{\frac {a+t}{x}}\right|\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
d
x
t
=
arcsin
x
a
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {dx}{t}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x
2
d
x
t
=
1
2
(
−
x
t
+
a
2
arcsin
x
a
)
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {x^{2}\;dx}{t}}={\frac {1}{2}}\left(-xt+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
t
d
x
=
1
2
(
x
t
−
sgn
x
cosh
−
1
|
x
a
|
)
(untuk
|
x
|
≥
|
a
|
)
{\displaystyle \int t\;dx={\frac {1}{2}}\left(xt-\operatorname {sgn} x\,\cosh ^{-1}\left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(untuk }}|x|\geq |a|{\mbox{)}}}
R
=
a
x
2
+
b
x
+
c
{\displaystyle R={\sqrt {ax^{2}+bx+c}}}
∫
d
x
R
=
1
a
ln
|
2
a
R
+
2
a
x
+
b
|
(untuk
a
>
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|\qquad {\mbox{(untuk }}a>0{\mbox{)}}}
∫
d
x
R
=
1
a
sinh
−
1
2
a
x
+
b
4
a
c
−
b
2
(untuk
a
>
0
,
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\,\sinh ^{-1}{\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(untuk }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫
d
x
R
=
1
a
ln
|
2
a
x
+
b
|
(untuk
a
>
0
,
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(untuk }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}
∫
d
x
R
=
−
1
−
a
arcsin
2
a
x
+
b
b
2
−
4
a
c
(untuk
a
<
0
,
4
a
c
−
b
2
<
0
,
|
2
a
x
+
b
|
<
b
2
−
4
a
c
)
{\displaystyle \int {\frac {dx}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(untuk }}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{, }}\left|2ax+b\right|<{\sqrt {b^{2}-4ac}}{\mbox{)}}}
∫
d
x
R
3
=
4
a
x
+
2
b
(
4
a
c
−
b
2
)
R
{\displaystyle \int {\frac {dx}{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 {dx}{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 {dx}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left({\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {dx}{R^{2n-1}}}\right)}
∫
x
R
d
x
=
R
a
−
b
2
a
∫
d
x
R
{\displaystyle \int {\frac {x}{R}}\;dx={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {dx}{R}}}
∫
x
R
3
d
x
=
−
2
b
x
+
4
c
(
4
a
c
−
b
2
)
R
{\displaystyle \int {\frac {x}{R^{3}}}\;dx=-{\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}}}\;dx=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{2n+1}}}}
∫
d
x
x
R
=
−
1
c
ln
(
2
c
R
+
b
x
+
2
c
x
)
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right)}
∫
d
x
x
R
=
−
1
c
sinh
−
1
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\sinh ^{-1}\left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}
S
=
a
x
+
b
{\displaystyle S={\sqrt {ax+b}}}
∫
d
x
x
a
x
+
b
=
−
2
b
tanh
−
1
a
x
+
b
b
{\displaystyle \int {\frac {dx}{x{\sqrt {ax+b}}}}\,=\,{\frac {-2}{\sqrt {b}}}\tanh ^{-1}{\sqrt {\frac {ax+b}{b}}}}
∫
a
x
+
b
x
d
x
=
2
(
a
x
+
b
−
b
tanh
−
1
a
x
+
b
b
)
{\displaystyle \int {\frac {\sqrt {ax+b}}{x}}\,dx\;=\;2\left({\sqrt {ax+b}}-{\sqrt {b}}\tanh ^{-1}{\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)}
= Referensi
=
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 1972, Dover: New York. (See chapter 3.)
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