[ تعرٌف على ] قائمة تكاملات الدوال غير الكسرية # اخر تحديث اليوم 2024-04-27

تم النشر اليوم 2024-04-27 | قائمة تكاملات الدوال غير الكسرية

توابع تحتوي
R 1 / 2
=
a
x
+
b
{\displaystyle R^{1/2}={\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 (
x n
+
1
a
x
+
b
+
b x n
a
x
+
b
−
n
b
∫ x n
−
1 a
x
+
b
) {\displaystyle \int {\frac {x^{n}}{\sqrt {ax+b}}}\,dx\;=\;{\frac {2}{a}}\left(x^{n+1}{\sqrt {ax+b}}+bx^{n}{\sqrt {ax+b}}-nb\int {\frac {x^{n-1}}{\sqrt {ax+b}}}\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)}

توابع تحتوي
R 1 / 2
=
a x 2
+
b
x
+
c
{\displaystyle R^{1/2}={\sqrt {ax^{2}+bx+c}}}

∫ d
x
a x 2
+
b
x
+
c =
1 a ln
⁡ | 2
a
R
+
2
a
x
+
b |
(for
a
>
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {aR}}+2ax+b\right|\qquad {\mbox{(for }}a>0{\mbox{)}}}
∫ d
x
a x 2
+
b
x
+
c =
1 a sinh −
1
⁡ 2
a
x
+
b
4
a
c
− b 2
(for
a
>
0
,
4
a
c
− b 2
>
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\,\sinh ^{-1}{\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫ d
x
a x 2
+
b
x
+
c =
1 a ln
⁡ | 2
a
x
+
b |
(for
a
>
0
,
4
a
c
− b 2
=
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}
∫ d
x
a x 2
+
b
x
+
c =
−
1 −
a arcsin
⁡ 2
a
x
+
b b 2
−
4
a
c
(for
a
<
0
,
4
a
c
− b 2
<
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(for }}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{)}}}
∫ d
x
(
a x 2
+
b
x
+
c ) 3 = 4
a
x
+
2
b
(
4
a
c
− b 2
)
R {\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{3}}}}={\frac {4ax+2b}{(4ac-b^{2}){\sqrt {R}}}}}
∫ d
x
(
a x 2
+
b
x
+
c ) 5 = 4
a
x
+
2
b
3
(
4
a
c
− b 2
)
R
( 1
R
+ 8
a
4
a
c
− b 2
) {\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{5}}}}={\frac {4ax+2b}{3(4ac-b^{2}){\sqrt {R}}}}\left({\frac {1}{R}}+{\frac {8a}{4ac-b^{2}}}\right)}
∫ d
x
(
a x 2
+
b
x
+
c ) 2
n
+
1 = 4
a
x
+
2
b
(
2
n
−
1
)
(
4
a
c
− b 2
) R (
2
n
−
1
) / 2 + 8
a
(
n
−
1
)
(
2
n
−
1
)
(
4
a
c
− b 2
) ∫ d
x
R (
2
n
−
1
) / 2
{\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}={\frac {4ax+2b}{(2n-1)(4ac-b^{2})R^{(2n-1)/2}}}+{\frac {8a(n-1)}{(2n-1)(4ac-b^{2})}}\int {\frac {dx}{R^{(2n-1)/2}}}}
∫ x d
x
a x 2
+
b
x
+
c = R a
−
b 2
a ∫ d
x
R {\displaystyle \int {\frac {x\;dx}{\sqrt {ax^{2}+bx+c}}}={\frac {\sqrt {R}}{a}}-{\frac {b}{2a}}\int {\frac {dx}{\sqrt {R}}}}
∫ x d
x
(
a x 2
+
b
x
+
c ) 3 =
− 2
b
x
+
4
c
(
4
a
c
− b 2
)
R {\displaystyle \int {\frac {x\;dx}{\sqrt {(ax^{2}+bx+c)^{3}}}}=-{\frac {2bx+4c}{(4ac-b^{2}){\sqrt {R}}}}}
∫ x d
x
(
a x 2
+
b
x
+
c ) 2
n
+
1 =
−
1 (
2
n
−
1
)
a R (
2
n
−
1
) / 2 −
b 2
a ∫ d
x
R (
2
n
+
1
) / 2
{\displaystyle \int {\frac {x\;dx}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}=-{\frac {1}{(2n-1)aR^{(2n-1)/2}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{(2n+1)/2}}}}
∫ d
x
x
a x 2
+
b
x
+
c =
−
1 c ln
⁡ ( 2
c
R
+
b
x
+
2
c x
) {\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {cR}}+bx+2c}{x}}\right)}
∫ d
x
x
a x 2
+
b
x
+
c =
−
1 c
sinh −
1
⁡ ( b
x
+
2
c | x | 4
a
c
− b 2 ) {\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\sinh ^{-1}\left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}

توابع تحتوي t
= a 2
− x 2
{\displaystyle t={\sqrt {a^{2}-x^{2}}}}

∫
t d
x
=
1
2 ( x
t
+ a 2 sin −
1
⁡
x
a )
( | x | ≤ | a | )
{\displaystyle \int t\;dx={\frac {1}{2}}\left(xt+a^{2}\sin ^{-1}{\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
= sin −
1
⁡
x
a ( | x | ≤ | a | )
{\displaystyle \int {\frac {dx}{t}}=\sin ^{-1}{\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x 2 d
x t
=
−
x
2
t
+ a 2
2 sin −
1
⁡
x
a ( | x | ≤ | a | )
{\displaystyle \int {\frac {x^{2}\;dx}{t}}=-{\frac {x}{2}}t+{\frac {a^{2}}{2}}\sin ^{-1}{\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
t d
x
=
1
2 ( x
t
−
sgn
⁡
x
cosh −
1
⁡ |
x
a
|
)
(for | x | ≥ | a | )
{\displaystyle \int t\;dx={\frac {1}{2}}\left(xt-\operatorname {sgn} x\,\cosh ^{-1}\left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(for }}|x|\geq |a|{\mbox{)}}}

توابع تحتوي s
= x 2
− a 2
{\displaystyle s={\sqrt {x^{2}-a^{2}}}}

بفرض أن ( x 2
> a 2
)
{\displaystyle (x^{2}>a^{2})} ، من أجل ( x 2
< a 2
)
{\displaystyle (x^{2} ∫
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|} لاحظ أن 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)} ، حيث يجب أخذ القيمة الإيجابية لـ
cosh −
1
⁡ |
x
a
| {\displaystyle \cosh ^{-1}\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
ln
⁡ | x
+
s a
| {\displaystyle \int {\frac {x^{4}\;dx}{s^{3}}}={\frac {xs}{2}}-{\frac {a^{2}x}{s}}+{\frac {3}{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]}

توابع تحتوي r
= a 2
+ x 2
{\displaystyle r={\sqrt {a^{2}+x^{2}}}}

∫
r d
x
=
1
2 ( x
r
+ a 2 ln
⁡ ( x
+
r a
)
) {\displaystyle \int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left({\frac {x+r}{a}}\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 a
) {\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {1}{8}}3a^{2}xr+{\frac {3}{8}}a^{4}\ln \left({\frac {x+r}{a}}\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 5
ln
⁡ ( x
+
r a
) {\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^{5}\ln \left({\frac {x+r}{a}}\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 a
) {\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left({\frac {x+r}{a}}\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 a
) {\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({\frac {x+r}{a}}\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 3
7
−
a 2 r 5 5
{\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{3}}{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 a
) {\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({\frac {x+r}{a}}\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 a
) {\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({\frac {x+r}{a}}\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 3 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^{3}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|}
∫ x d
x r
=
r
{\displaystyle \int {\frac {x\,dx}{r}}=r}
∫
x 2 d
x r
=
x
2
r
− a 2
2
sinh −
1
⁡
a
x
=
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 {a}{x}}={\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|}

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