Tích phân chứa r = √a 2 + x 2
sửa
∫
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
2
r
2
n
+
3
2
n
+
3
{\displaystyle \int x^{3}r^{2n+1}\,dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{2}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|}
Tích phân chứa s = √x 2 − a 2
sửa
Giả định x 2 > a 2 (đối với trường hợp x 2 < a 2 , xem mục sau):
∫
s
d
x
=
1
2
(
x
s
−
a
2
ln
|
x
+
s
|
)
{\displaystyle \int s\,dx={\frac {1}{2}}\left(xs-a^{2}\ln \left|x+s\right|\right)}
∫
x
s
d
x
=
1
3
s
3
{\displaystyle \int xs\,dx={\frac {1}{3}}s^{3}}
∫
s
d
x
x
=
s
−
|
a
|
arccos
|
a
x
|
{\displaystyle \int {\frac {s\,dx}{x}}=s-|a|\arccos \left|{\frac {a}{x}}\right|}
∫
d
x
s
=
ln
|
x
+
s
a
|
.
{\displaystyle \int {\frac {dx}{s}}=\ln \left|{\frac {x+s}{a}}\right|\,.}
ln
|
x
+
s
a
|
=
sgn
(
x
)
arcosh
|
x
a
|
=
1
2
ln
(
x
+
s
x
−
s
)
,
{\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\operatorname {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|}
∫
d
x
x
s
=
1
a
arcsec
|
x
a
|
{\displaystyle \int {\frac {dx}{xs}}={\frac {1}{a}}\operatorname {arcsec} \left|{\frac {x}{a}}\right|}
∫
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]}
Tích phân chứa u = √a 2 − x 2
sửa
∫
u
d
x
=
1
2
(
x
u
+
a
2
arcsin
x
a
)
(
|
x
|
≤
|
a
|
)
{\displaystyle \int u\,dx={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x
u
d
x
=
−
1
3
u
3
(
|
x
|
≤
|
a
|
)
{\displaystyle \int xu\,dx=-{\frac {1}{3}}u^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
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\,dx=-{\frac {x}{4}}u^{3}+{\frac {a^{2}}{8}}(xu+a^{2}\arcsin {\frac {x}{a}})\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
u
d
x
x
=
u
−
a
ln
|
a
+
u
x
|
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {u\,dx}{x}}=u-a\ln \left|{\frac {a+u}{x}}\right|\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
d
x
u
=
arcsin
x
a
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {dx}{u}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x
2
d
x
u
=
1
2
(
−
x
u
+
a
2
arcsin
x
a
)
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {x^{2}\,dx}{u}}={\frac {1}{2}}\left(-xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
u
d
x
=
1
2
(
x
u
−
sgn
x
arcosh
|
x
a
|
)
(với
|
x
|
≥
|
a
|
)
{\displaystyle \int u\,dx={\frac {1}{2}}\left(xu-\operatorname {sgn} x\,\operatorname {arcosh} \left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(với }}|x|\geq |a|{\mbox{)}}}
∫
x
u
d
x
=
−
u
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {x}{u}}\,dx=-u\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
Tích phân chứa R = √ax 2 + bx + c
sửa
Giả định rằng (ax 2 + bx + c ) không thể chuyển về dạng (px + q )2 với p và q là hằng số.
∫
d
x
R
=
1
a
ln
|
2
a
R
+
2
a
x
+
b
|
(với
a
>
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|\qquad {\mbox{(với }}a>0{\mbox{)}}}
∫
d
x
R
=
1
a
arsinh
2
a
x
+
b
4
a
c
−
b
2
(với
a
>
0
,
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\,\operatorname {arsinh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(với }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫
d
x
R
=
1
a
ln
|
2
a
x
+
b
|
(với
a
>
0
,
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(với }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}
∫
d
x
R
=
−
1
−
a
arcsin
2
a
x
+
b
b
2
−
4
a
c
(với
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{(với }}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
|
,
c
>
0
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left|{\frac {2{\sqrt {c}}R+bx+2c}{x}}\right|,~c>0}
∫
d
x
x
R
=
−
1
c
arsinh
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
,
c
<
0
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right),~c<0}
∫
d
x
x
R
=
1
−
c
arcsin
(
b
x
+
2
c
|
x
|
b
2
−
4
a
c
)
,
c
<
0
,
b
2
−
4
a
c
>
0
{\displaystyle \int {\frac {dx}{xR}}={\frac {1}{\sqrt {-c}}}\operatorname {arcsin} \left({\frac {bx+2c}{|x|{\sqrt {b^{2}-4ac}}}}\right),~c<0,b^{2}-4ac>0}
∫
d
x
x
R
=
−
2
b
x
(
a
x
2
+
b
x
)
,
c
=
0
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {2}{bx}}\left({\sqrt {ax^{2}+bx}}\right),~c=0}
∫
x
2
R
d
x
=
2
a
x
−
3
b
4
a
2
R
+
3
b
2
−
4
a
c
8
a
2
∫
d
x
R
{\displaystyle \int {\frac {x^{2}}{R}}\,dx={\frac {2ax-3b}{4a^{2}}}R+{\frac {3b^{2}-4ac}{8a^{2}}}\int {\frac {dx}{R}}}
∫
d
x
x
2
R
=
−
R
c
x
−
b
2
c
∫
d
x
x
R
{\displaystyle \int {\frac {dx}{x^{2}R}}=-{\frac {R}{cx}}-{\frac {b}{2c}}\int {\frac {dx}{xR}}}
∫
R
d
x
=
2
a
x
+
b
4
a
R
+
4
a
c
−
b
2
8
a
∫
d
x
R
{\displaystyle \int R\,dx={\frac {2ax+b}{4a}}R+{\frac {4ac-b^{2}}{8a}}\int {\frac {dx}{R}}}
∫
x
R
d
x
=
R
3
3
a
−
b
(
2
a
x
+
b
)
8
a
2
R
−
b
(
4
a
c
−
b
2
)
16
a
2
∫
d
x
R
{\displaystyle \int xR\,dx={\frac {R^{3}}{3a}}-{\frac {b(2ax+b)}{8a^{2}}}R-{\frac {b(4ac-b^{2})}{16a^{2}}}\int {\frac {dx}{R}}}
∫
x
2
R
d
x
=
6
a
x
−
5
b
24
a
2
R
3
+
5
b
2
−
4
a
c
16
a
2
∫
R
d
x
{\displaystyle \int x^{2}R\,dx={\frac {6ax-5b}{24a^{2}}}R^{3}+{\frac {5b^{2}-4ac}{16a^{2}}}\int R\,dx}
∫
R
x
d
x
=
R
+
b
2
∫
d
x
R
+
c
∫
d
x
x
R
{\displaystyle \int {\frac {R}{x}}\,dx=R+{\frac {b}{2}}\int {\frac {dx}{R}}+c\int {\frac {dx}{xR}}}
∫
R
x
2
d
x
=
−
R
x
+
a
∫
d
x
R
+
b
2
∫
d
x
x
R
{\displaystyle \int {\frac {R}{x^{2}}}\,dx=-{\frac {R}{x}}+a\int {\frac {dx}{R}}+{\frac {b}{2}}\int {\frac {dx}{xR}}}
∫
x
2
d
x
R
3
=
(
2
b
2
−
4
a
c
)
x
+
2
b
c
a
(
4
a
c
−
b
2
)
R
+
1
a
∫
d
x
R
{\displaystyle \int {\frac {x^{2}\,dx}{R^{3}}}={\frac {(2b^{2}-4ac)x+2bc}{a(4ac-b^{2})R}}+{\frac {1}{a}}\int {\frac {dx}{R}}}
Tích phân chứa S = √ax + b
sửa
∫
S
d
x
=
2
S
3
3
a
{\displaystyle \int S\,dx={\frac {2S^{3}}{3a}}}
∫
d
x
S
=
2
S
a
{\displaystyle \int {\frac {dx}{S}}={\frac {2S}{a}}}
∫
d
x
x
S
=
{
−
2
b
arcoth
(
S
b
)
(với
b
>
0
,
a
x
>
0
)
−
2
b
artanh
(
S
b
)
(với
b
>
0
,
a
x
<
0
)
2
−
b
arctan
(
S
−
b
)
(với
b
<
0
)
{\displaystyle \int {\frac {dx}{xS}}={\begin{cases}-{\dfrac {2}{\sqrt {b}}}\operatorname {arcoth} \left({\dfrac {S}{\sqrt {b}}}\right)&{\mbox{(với }}b>0,\quad ax>0{\mbox{)}}\\-{\dfrac {2}{\sqrt {b}}}\operatorname {artanh} \left({\dfrac {S}{\sqrt {b}}}\right)&{\mbox{(với }}b>0,\quad ax<0{\mbox{)}}\\{\dfrac {2}{\sqrt {-b}}}\arctan \left({\dfrac {S}{\sqrt {-b}}}\right)&{\mbox{(với }}b<0{\mbox{)}}\\\end{cases}}}
∫
S
x
d
x
=
{
2
(
S
−
b
arcoth
(
S
b
)
)
(với
b
>
0
,
a
x
>
0
)
2
(
S
−
b
artanh
(
S
b
)
)
(với
b
>
0
,
a
x
<
0
)
2
(
S
−
−
b
arctan
(
S
−
b
)
)
(với
b
<
0
)
{\displaystyle \int {\frac {S}{x}}\,dx={\begin{cases}2\left(S-{\sqrt {b}}\,\operatorname {arcoth} \left({\dfrac {S}{\sqrt {b}}}\right)\right)&{\mbox{(với }}b>0,\quad ax>0{\mbox{)}}\\2\left(S-{\sqrt {b}}\,\operatorname {artanh} \left({\dfrac {S}{\sqrt {b}}}\right)\right)&{\mbox{(với }}b>0,\quad ax<0{\mbox{)}}\\2\left(S-{\sqrt {-b}}\arctan \left({\dfrac {S}{\sqrt {-b}}}\right)\right)&{\mbox{(với }}b<0{\mbox{)}}\\\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}}\,dx={\frac {2}{a(2n+1)}}\left(x^{n}S-bn\int {\frac {x^{n-1}}{S}}\,dx\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\,dx={\frac {2}{a(2n+3)}}\left(x^{n}S^{3}-nb\int x^{n-1}S\,dx\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}}\,dx=-{\frac {1}{b(n-1)}}\left({\frac {S}{x^{n-1}}}+\left(n-{\frac {3}{2}}\right)a\int {\frac {dx}{x^{n-1}S}}\right)}
Tham khảo
sửa
Abramowitz, Milton; Stegun, Irene A. biên tập (1972). “Chapter 3” . Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables . New York: Dover. tr. 12–13. Gradshteyn, Izrail Solomonovich; Ryzhik, Iosif Moiseevich; Geronimus, Yuri Veniaminovich; Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo (biên tập). Table of Integrals, Series, and Products (bằng tiếng Anh). Scripta Technica, Inc. biên dịch (ấn bản 8). Academic Press, Inc. ISBN 978-0-12-384933-5 LCCN 2014010276 . Peirce, Benjamin Osgood (1929) [1899]. “Chapter 3”. A Short Table of Integrals (ấn bản 3). Boston: Ginn and Co. tr. 16–30. 
![{\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/054a5959ce5e03cf279c1b29dff2ba014ac6dcde)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86843311de7fc72bc01f87742445f7c4b88899e9)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca32b3a8d7f9040840f5d1de3467129edff0d80b)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a98057cf3f3d6b7025114445c972bb6b7b7af9d)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cce4b87e7a47ce42042803038139f830afd5d37)
