Интегралы с корнем из a 2 + x 2
править
Интегралы с корнем из x 2 − a 2
править
Интегралы с корнем из a 2 − x 2
править
Здесь обозначено:
R
=
a
x
2
+
b
x
+
c
{\displaystyle R=ax^{2}+bx+c}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
ln
|
2
a
R
+
2
a
x
+
b
|
(
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{( }}a>0{\mbox{)}}}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
arsh
2
a
x
+
b
4
a
c
−
b
2
(
a
>
0
,
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\,\operatorname {arsh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{( }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
ln
|
2
a
x
+
b
|
(
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{( }}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
(
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{( }}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
arsh
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
{\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}
∫
a
x
2
+
b
x
+
c
d
x
=
(
x
2
+
b
4
a
)
a
x
2
+
b
x
+
c
+
(
c
2
−
b
2
8
a
)
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \int {\sqrt {ax^{2}+bx+c}}\,dx=({\frac {x}{2}}+{\frac {b}{4a}}){\sqrt {ax^{2}+bx+c}}+({\frac {c}{2}}-{\frac {b^{2}}{8a}})\int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}\qquad }
, где последний интеграл находится в зависимости от параметров a,b и c (см. выше)
∫
d
x
x
a
x
+
b
=
−
2
b
arth
a
x
+
b
b
{\displaystyle \int {\frac {dx}{x{\sqrt {ax+b}}}}\,=\,{\frac {-2}{\sqrt {b}}}\operatorname {arth} {\sqrt {\frac {ax+b}{b}}}}
∫
a
x
+
b
x
d
x
=
2
(
a
x
+
b
−
b
arth
a
x
+
b
b
)
{\displaystyle \int {\frac {\sqrt {ax+b}}{x}}\,dx\;=\;2\left({\sqrt {ax+b}}-{\sqrt {b}}\operatorname {arth} {\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)}
Книги
Таблицы интегралов
Вычисление интегралов