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積分表
維基媒體列表條目
語言
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編輯
目次
1
含有'"`UNIQ--postMath-00000001-QINU`"'的積分
2
含有'"`UNIQ--postMath-00000008-QINU`"'的積分
3
含有'"`UNIQ--postMath-00000012-QINU`"'的積分
4
含有 '"`UNIQ--postMath-00000015-QINU`"'的積分
5
含有 '"`UNIQ--postMath-00000017-QINU`"'的積分
6
含有 '"`UNIQ--postMath-00000019-QINU`"'的積分
7
含有'"`UNIQ--postMath-00000022-QINU`"'的積分
8
含有'"`UNIQ--postMath-00000024-QINU`"'的積分
9
含有'"`UNIQ--postMath-0000002D-QINU`"'的積分
10
含有三角函數的積分
11
含有反三角函數的積分
12
含有指數函數的積分
13
含有對數函數的積分
14
含有雙曲函數的積分
15
定積分
由於列表比較長,
積分表
被分為以下幾個部分:
有理函數積分表
無理函數積分表
指數函數積分表
對數函數積分表
高斯函數積分表
三角函數積分表
反三角函數積分表
雙曲函數積分表
反雙曲函數積分表
含有
a
x
+
b
{\displaystyle ax+b}
的積分
編輯
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
{\displaystyle \int \ (ax+b)^{n}{\mbox{d}}x={\frac {(ax+b)^{n+1}}{a(n+1)}}+C}
∫
1
a
x
+
b
d
x
=
1
a
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {1}{ax+b}}{\mbox{d}}x={\frac {1}{a}}\ln \left|ax+b\right|+C}
∫
x
a
x
+
b
d
x
=
1
a
2
(
a
x
+
b
−
b
ln
|
a
x
+
b
|
)
+
C
{\displaystyle \int {\frac {x}{ax+b}}{\mbox{d}}x={\frac {1}{a^{2}}}(ax+b-b\ln \left|ax+b\right|)+C}
∫
x
2
a
x
+
b
d
x
=
1
2
a
3
[
(
a
x
+
b
)
2
−
4
b
(
a
x
+
b
)
+
2
b
2
ln
|
a
x
+
b
|
]
+
C
{\displaystyle \int {\frac {x^{2}}{ax+b}}{\mbox{d}}x={\frac {1}{2a^{3}}}\left[(ax+b)^{2}-4b(ax+b)+2b^{2}\ln \left|ax+b\right|\right]+C}
∫
1
x
(
a
x
+
b
)
d
x
=
−
1
b
ln
|
a
x
+
b
x
|
+
C
{\displaystyle \int {\frac {1}{x(ax+b)}}{\mbox{d}}x=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C}
∫
1
x
2
(
a
x
+
b
)
d
x
=
a
b
2
ln
|
a
x
+
b
x
|
−
1
b
x
+
C
{\displaystyle \int {\frac {1}{x^{2}(ax+b)}}{\mbox{d}}x={\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|-{\frac {1}{bx}}+C}
含有
a
+
b
x
{\displaystyle {\sqrt {a+bx}}}
的積分
編輯
∫
x
a
+
b
x
d
x
=
2
15
b
2
(
3
b
x
−
2
a
)
(
a
+
b
x
)
3
2
+
C
{\displaystyle \int x{\sqrt {a+bx}}{\mbox{d}}x={\frac {2}{15b^{2}}}(3bx-2a)(a+bx)^{\frac {3}{2}}+C}
∫
x
2
a
+
b
x
d
x
=
2
105
b
3
(
15
b
2
x
2
−
12
a
b
x
+
8
a
2
)
(
a
+
b
x
)
3
2
+
C
{\displaystyle \int x^{2}{\sqrt {a+bx}}{\mbox{d}}x={\frac {2}{105b^{3}}}(15b^{2}x^{2}-12abx+8a^{2})(a+bx)^{\frac {3}{2}}+C}
∫
x
n
a
+
b
x
d
x
=
2
b
(
2
n
+
3
)
x
n
(
a
+
b
x
)
3
2
−
2
n
a
b
(
2
n
+
3
)
∫
x
n
−
1
a
+
b
x
d
x
{\displaystyle \int x^{n}{\sqrt {a+bx}}{\mbox{d}}x={\frac {2}{b(2n+3)}}x^{n}(a+bx)^{\frac {3}{2}}-{\frac {2na}{b(2n+3)}}\int x^{n-1}{\sqrt {a+bx}}{\mbox{d}}x}
∫
a
+
b
x
x
d
x
=
2
a
+
b
x
+
a
∫
1
x
a
+
b
x
d
x
{\displaystyle \int {\frac {\sqrt {a+bx}}{x}}{\mbox{d}}x=2{\sqrt {a+bx}}+a\int {\frac {1}{x{\sqrt {a+bx}}}}{\mbox{d}}x}
∫
a
+
b
x
x
n
d
x
=
−
1
a
(
n
−
1
)
(
a
+
b
x
)
3
2
x
n
−
1
−
(
2
n
−
5
)
b
2
a
(
n
−
1
)
∫
a
+
b
x
x
n
−
1
d
x
,
n
≠
1
{\displaystyle \int {\frac {\sqrt {a+bx}}{x^{n}}}{\mbox{d}}x={\frac {-1}{a(n-1)}}{\frac {(a+bx)^{\frac {3}{2}}}{x^{n-1}}}-{\frac {(2n-5)b}{2a(n-1)}}\int {\frac {\sqrt {a+bx}}{x^{n-1}}}{\mbox{d}}x,n\neq 1}
∫
1
x
a
+
b
x
d
x
=
1
a
ln
(
a
+
b
x
−
a
a
+
b
x
+
a
)
+
C
,
a
>
0
{\displaystyle \int {\frac {1}{x{\sqrt {a+bx}}}}{\mbox{d}}x={\frac {1}{\sqrt {a}}}\ln \left({\frac {{\sqrt {a+bx}}-{\sqrt {a}}}{{\sqrt {a+bx}}+{\sqrt {a}}}}\right)+C,a>0}
=
2
−
a
arctan
a
+
b
x
−
a
+
C
,
a
<
0
{\displaystyle ={\frac {2}{\sqrt {-a}}}\arctan {\sqrt {\frac {a+bx}{-a}}}+C,a<0}
∫
x
a
+
b
x
d
x
=
2
(
a
+
b
x
)
3
2
3
b
2
−
(
2
a
)
a
+
b
x
b
2
{\displaystyle \int {\frac {x}{\sqrt {a+bx}}}{\mbox{d}}x={\frac {2(a+bx)^{\frac {3}{2}}}{3b^{2}}}-{\frac {(2a){\sqrt {a+bx}}}{b^{2}}}}
∫
1
x
n
a
+
b
x
d
x
=
−
1
a
(
n
−
1
)
a
+
b
x
x
n
−
1
−
(
2
n
−
3
)
b
2
a
(
n
−
1
)
∫
1
x
n
−
1
a
+
b
x
d
x
,
n
≠
1
{\displaystyle \int {\frac {1}{x^{n}{\sqrt {a+bx}}}}{\mbox{d}}x={\frac {-1}{a(n-1)}}{\frac {\sqrt {a+bx}}{x^{n-1}}}-{\frac {(2n-3)b}{2a(n-1)}}\int {\frac {1}{x^{n-1}}}{\sqrt {a+bx}}{\mbox{d}}x,n\neq 1}
含有
x
2
±
α
2
{\displaystyle x^{2}\pm \alpha ^{2}}
的積分
編輯
∫
1
x
2
+
α
2
d
x
=
arctan
x
α
α
+
C
{\displaystyle \int {\frac {1}{x^{2}+\alpha ^{2}}}{\mbox{d}}x={\frac {\arctan {\dfrac {x}{\alpha }}}{\alpha }}+C}
∫
1
±
x
2
∓
α
2
d
x
=
ln
(
x
∓
α
±
x
+
α
)
2
α
+
C
{\displaystyle \int {\frac {1}{\pm x^{2}\mp \alpha ^{2}}}{\mbox{d}}x={\frac {\ln \left({\dfrac {x\mp \alpha }{\pm x+\alpha }}\right)}{2\alpha }}+C}
含有
a
x
2
+
b
{\displaystyle {ax^{2}+b}}
的積分
編輯
∫
1
a
x
2
+
b
d
x
=
1
a
b
arctan
a
x
b
+
C
{\displaystyle \int {\frac {1}{ax^{2}+b}}{\mbox{d}}x={\frac {1}{\sqrt {ab}}}\arctan {\frac {{\sqrt {a}}x}{\sqrt {b}}}+C}
含有
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
的積分
編輯
∫
(
a
x
2
+
b
x
+
c
)
d
x
=
a
x
3
3
+
b
x
2
2
+
c
x
+
C
{\displaystyle \int (ax^{2}+bx+c){\mbox{d}}x={\frac {ax^{3}}{3}}+{\frac {bx^{2}}{2}}+cx+C}
含有
a
2
+
x
2
(
a
>
0
)
{\displaystyle {\sqrt {a^{2}+x^{2}}}\qquad (a>0)}
的積分
編輯
∫
a
2
+
x
2
d
x
=
1
2
x
a
2
+
x
2
+
1
2
a
2
ln
(
x
+
a
2
+
x
2
)
+
C
{\displaystyle \int {\sqrt {a^{2}+x^{2}}}{\mbox{d}}x={\frac {1}{2}}x{\sqrt {a^{2}+x^{2}}}+{\frac {1}{2}}a^{2}\ln \left(x+{\sqrt {a^{2}+x^{2}}}\right)+C}
∫
x
2
a
2
+
x
2
d
x
=
1
8
x
(
a
2
+
2
x
2
)
a
2
+
x
2
−
1
8
a
4
ln
(
x
+
a
2
+
x
2
)
+
C
{\displaystyle \int x^{2}{\sqrt {a^{2}+x^{2}}}{\mbox{d}}x={\frac {1}{8}}x(a^{2}+2x^{2}){\sqrt {a^{2}+x^{2}}}-{\frac {1}{8}}a^{4}\ln \left(x+{\sqrt {a^{2}+x^{2}}}\right)+C}
∫
a
2
+
x
2
x
d
x
=
a
2
+
x
2
−
a
ln
(
a
+
a
2
+
x
2
x
)
+
C
{\displaystyle \int {\frac {\sqrt {a^{2}+x^{2}}}{x}}{\mbox{d}}x={\sqrt {a^{2}+x^{2}}}-a\ln \left({\frac {a+{\sqrt {a^{2}+x^{2}}}}{x}}\right)+C}
∫
a
2
+
x
2
x
2
d
x
=
ln
(
x
+
a
2
+
x
2
)
−
a
2
+
x
2
x
+
C
{\displaystyle \int {\frac {\sqrt {a^{2}+x^{2}}}{x^{2}}}{\mbox{d}}x=\ln \left(x+{\sqrt {a^{2}+x^{2}}}\right)-{\frac {\sqrt {a^{2}+x^{2}}}{x}}+C}
∫
1
a
2
+
x
2
d
x
=
ln
(
x
+
a
2
+
x
2
)
+
C
{\displaystyle \int {\frac {1}{\sqrt {a^{2}+x^{2}}}}{\mbox{d}}x=\ln \left(x+{\sqrt {a^{2}+x^{2}}}\right)+C}
∫
x
2
a
2
+
x
2
d
x
=
1
2
x
a
2
+
x
2
−
1
2
a
2
ln
(
a
2
+
x
2
+
x
)
+
C
{\displaystyle \int {\frac {x^{2}}{\sqrt {a^{2}+x^{2}}}}{\mbox{d}}x={\frac {1}{2}}x{\sqrt {a^{2}+x^{2}}}-{\frac {1}{2}}a^{2}\ln \left({\sqrt {a^{2}+x^{2}}}+x\right)+C}
∫
1
x
a
2
+
x
2
d
x
=
1
a
ln
(
x
a
+
a
2
+
x
2
)
+
C
{\displaystyle \int {\frac {1}{x{\sqrt {a^{2}+x^{2}}}}}{\mbox{d}}x={\frac {1}{a}}\ln \left({\frac {x}{a+{\sqrt {a^{2}+x^{2}}}}}\right)+C}
∫
1
x
2
a
2
+
x
2
d
x
=
−
a
2
+
x
2
a
2
x
+
C
{\displaystyle \int {\frac {1}{x^{2}{\sqrt {a^{2}+x^{2}}}}}{\mbox{d}}x=-{\frac {\sqrt {a^{2}+x^{2}}}{a^{2}x}}+C}
含有
x
2
−
a
2
(
x
2
>
a
2
)
{\displaystyle {\sqrt {x^{2}-a^{2}}}\qquad {(x^{2}>a^{2})}}
的積分
編輯
∫
1
x
2
−
a
2
d
x
=
l
n
|
x
+
x
2
−
a
2
|
+
C
{\displaystyle \int {\frac {1}{\sqrt {x^{2}-a^{2}}}}{\mbox{d}}x=ln|x+{\sqrt {x^{2}-a^{2}}}|+C}
含有
a
2
−
x
2
(
a
2
>
x
2
)
{\displaystyle {\sqrt {a^{2}-x^{2}}}\qquad (a^{2}>x^{2})}
的積分
編輯
∫
a
2
−
x
2
d
x
=
1
2
x
a
2
−
x
2
+
a
2
2
arcsin
x
a
+
C
{\displaystyle \int {\sqrt {a^{2}-x^{2}}}{\mbox{d}}x={\frac {1}{2}}x{\sqrt {a^{2}-x^{2}}}+{\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+C}
∫
1
a
2
−
x
2
d
x
=
arcsin
x
a
+
C
=
−
arccos
x
a
+
C
{\displaystyle \int {\frac {1}{\sqrt {a^{2}-x^{2}}}}{\mbox{d}}x=\arcsin {\frac {x}{a}}+C=-\arccos {\frac {x}{a}}+C}
∫
x
2
a
2
−
x
2
d
x
=
1
8
x
(
2
x
2
−
a
2
)
a
2
−
x
2
+
1
8
a
4
arcsin
x
a
+
C
{\displaystyle \int x^{2}{\sqrt {a^{2}-x^{2}}}{\mbox{d}}x={\frac {1}{8}}x(2x^{2}-a^{2}){\sqrt {a^{2}-x^{2}}}+{\frac {1}{8}}a^{4}\arcsin {\frac {x}{a}}+C}
∫
a
2
−
x
2
x
d
x
=
a
2
−
x
2
−
a
ln
(
a
+
a
2
−
x
2
x
)
+
C
{\displaystyle \int {\frac {\sqrt {a^{2}-x^{2}}}{x}}{\mbox{d}}x={\sqrt {a^{2}-x^{2}}}-a\ln \left({\frac {a+{\sqrt {a^{2}-x^{2}}}}{x}}\right)+C}
∫
a
2
−
x
2
x
2
d
x
=
−
a
2
−
x
2
x
−
arcsin
x
a
+
C
{\displaystyle \int {\frac {\sqrt {a^{2}-x^{2}}}{x^{2}}}{\mbox{d}}x=-{\frac {\sqrt {a^{2}-x^{2}}}{x}}-\arcsin {\frac {x}{a}}+C}
∫
1
x
a
2
−
x
2
d
x
=
−
1
a
ln
(
a
+
a
2
−
x
2
x
)
+
C
{\displaystyle \int {\frac {1}{x{\sqrt {a^{2}-x^{2}}}}}{\mbox{d}}x=-{\frac {1}{a}}\ln \left({\frac {a+{\sqrt {a^{2}-x^{2}}}}{x}}\right)+C}
∫
x
2
a
2
−
x
2
d
x
=
−
1
2
x
a
2
−
x
2
+
1
2
a
2
arcsin
x
a
+
C
{\displaystyle \int {\frac {x^{2}}{\sqrt {a^{2}-x^{2}}}}{\mbox{d}}x=-{\frac {1}{2}}x{\sqrt {a^{2}-x^{2}}}+{\frac {1}{2}}a^{2}\arcsin {\frac {x}{a}}+C}
∫
1
x
2
a
2
−
x
2
d
x
=
−
a
2
−
x
2
a
2
x
+
C
{\displaystyle \int {\frac {1}{x^{2}{\sqrt {a^{2}-x^{2}}}}}{\mbox{d}}x=-{\frac {\sqrt {a^{2}-x^{2}}}{a^{2}x}}+C}
含有
R
=
|
a
|
x
2
+
b
x
+
c
(
a
≠
0
)
{\displaystyle R={\sqrt {|a|x^{2}+bx+c}}\qquad (a\neq 0)}
的積分
編輯
∫
d
x
R
=
1
a
ln
(
2
a
R
+
2
a
x
+
b
)
(
for
a
>
0
)
{\displaystyle \int {\frac {{\mbox{d}}x}{R}}={\frac {1}{\sqrt {a}}}\ln \left(2{\sqrt {a}}R+2ax+b\right)\qquad ({\mbox{for }}a>0)}
∫
d
x
R
=
1
a
arsinh
2
a
x
+
b
4
a
c
−
b
2
(for
a
>
0
,
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {{\mbox{d}}x}{R}}={\frac {1}{\sqrt {a}}}\,\operatorname {arsinh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫
d
x
R
=
1
a
ln
|
2
a
x
+
b
|
(for
a
>
0
,
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {{\mbox{d}}x}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}
∫
d
x
R
=
−
1
−
a
arcsin
2
a
x
+
b
b
2
−
4
a
c
(for
a
<
0
,
4
a
c
−
b
2
<
0
,
(
2
a
x
+
b
)
<
b
2
−
4
a
c
)
{\displaystyle \int {\frac {{\mbox{d}}x}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(for }}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 {{\mbox{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 {{\mbox{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 {{\mbox{d}}x}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left[{\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {{\mbox{d}}x}{R^{2n-1}}}\right]}
∫
x
R
d
x
=
R
a
−
b
2
a
∫
d
x
R
{\displaystyle \int {\frac {x}{R}}\;{\mbox{d}}x={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {{\mbox{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}}}\;{\mbox{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}}}\;{\mbox{d}}x=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {{\mbox{d}}x}{R^{2n+1}}}}
∫
d
x
x
R
=
−
1
c
ln
(
2
c
R
+
b
x
+
2
c
x
)
{\displaystyle \int {\frac {{\mbox{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 {{\mbox{d}}x}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}
含有三角函數的積分
編輯
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos x{\mbox{d}}x=\sin x+C}
∫
sin
x
d
x
=
−
cos
x
+
C
{\displaystyle \int \sin x{\mbox{d}}x=-\cos x+C}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x{\mbox{d}}x=\tan x+C}
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int \csc ^{2}x{\mbox{d}}x=-\cot x+C}
∫
sec
x
tan
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec x\tan x{\mbox{d}}x=\sec x+C}
∫
csc
x
cot
x
d
x
=
−
csc
x
+
C
{\displaystyle \int \csc x\cot x{\mbox{d}}x=-\csc x+C}
∫
tan
x
d
x
=
−
ln
|
cos
x
|
+
C
=
ln
|
sec
x
|
+
C
{\displaystyle \int \tan x{\mbox{d}}x=-\ln {\left|\cos {x}\right|}+C=\ln {\left|\sec x\right|}+C}
∫
cot
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int \cot x{\mbox{d}}x=\ln {\left|\sin x\right|}+C}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
{\displaystyle \int \sec x{\mbox{d}}x=\ln {\left|\sec x+\tan x\right|}+C}
∫
csc
x
d
x
=
ln
|
csc
x
−
cot
x
|
+
C
=
ln
|
tan
x
−
sin
x
sin
x
tan
x
|
+
C
{\displaystyle \int \csc x{\mbox{d}}x=\ln {\left|\csc x-\cot x\right|}+C=\ln {\left|{\tan x-\sin x \over \sin x\tan x}\right|}+C}
∫
sin
n
x
d
x
=
−
1
n
sin
n
−
1
x
cos
x
+
n
−
1
n
∫
sin
n
−
2
x
d
x
+
C
∀
n
≥
2
{\displaystyle \int \sin ^{n}x{\mbox{d}}x=-{\frac {1}{n}}\sin ^{n-1}x\cos x+{\frac {n-1}{n}}\int \sin ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2}
∫
sin
2
x
d
x
=
x
2
−
sin
2
x
4
+
C
{\displaystyle \int \sin ^{2}x{\mbox{d}}x={\frac {x}{2}}-{\frac {\sin {2x}}{4}}+C}
∫
cos
n
x
d
x
=
1
n
cos
n
−
1
x
sin
x
+
n
−
1
n
∫
cos
n
−
2
x
d
x
+
C
∀
n
≥
2
{\displaystyle \int \cos ^{n}x{\mbox{d}}x={\frac {1}{n}}\cos ^{n-1}x\sin x+{\frac {n-1}{n}}\int \cos ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2}
∫
cos
2
x
d
x
=
x
2
+
sin
2
x
4
+
C
{\displaystyle \int \cos ^{2}x{\mbox{d}}x={\frac {x}{2}}+{\frac {\sin {2x}}{4}}+C}
∫
tan
n
x
d
x
=
1
n
−
1
tan
n
−
1
x
−
∫
tan
n
−
2
x
d
x
+
C
∀
n
≥
2
{\displaystyle \int \tan ^{n}x{\mbox{d}}x={\frac {1}{n-1}}\tan ^{n-1}x-\int \tan ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2}
∫
tan
2
x
d
x
=
tan
x
−
x
+
C
{\displaystyle \int \tan ^{2}x{\mbox{d}}x=\tan x-x+C}
∫
cot
n
x
d
x
=
−
1
n
−
1
cot
n
−
1
x
−
∫
cot
n
−
2
x
d
x
+
C
∀
n
≥
2
{\displaystyle \int \cot ^{n}x{\mbox{d}}x=-{\frac {1}{n-1}}\cot ^{n-1}x-\int \cot ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2}
∫
cot
2
x
d
x
=
−
cot
x
−
x
+
C
{\displaystyle \int \cot ^{2}x{\mbox{d}}x=-\cot x-x+C}
∫
sec
n
x
d
x
=
1
n
−
1
sec
n
−
2
x
tan
x
+
n
−
2
n
−
1
∫
sec
n
−
2
x
d
x
+
C
∀
n
≥
2
{\displaystyle \int \sec ^{n}x{\mbox{d}}x={\frac {1}{n-1}}\sec ^{n-2}x\tan x+{\frac {n-2}{n-1}}\int \sec ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2}
∫
csc
n
x
d
x
=
−
1
n
−
1
csc
n
−
2
x
cot
x
+
n
−
2
n
−
1
∫
csc
n
−
2
x
d
x
+
C
∀
n
≥
2
{\displaystyle \int \csc ^{n}x{\mbox{d}}x=-{\frac {1}{n-1}}\csc ^{n-2}x\cot x+{\frac {n-2}{n-1}}\int \csc ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2}
含有反三角函數的積分
編輯
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
C
{\displaystyle \int \arcsin x{\mbox{d}}x=x\arcsin x+{\sqrt {1-x^{2}}}+C}
∫
arccos
x
d
x
=
x
arccos
x
−
1
−
x
2
+
C
{\displaystyle \int \arccos x{\mbox{d}}x=x\arccos x-{\sqrt {1-x^{2}}}+C}
∫
arctan
x
d
x
=
x
arctan
x
−
1
2
ln
|
1
+
x
2
|
+
C
{\displaystyle \int \arctan {x}\,dx=x\arctan {x}-{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C}
∫
arccot
x
d
x
=
x
arccot
x
+
1
2
ln
|
1
+
x
2
|
+
C
{\displaystyle \int \operatorname {arccot} {x}\,dx=x\operatorname {arccot} {x}+{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C}
∫
arcsec
x
d
x
=
x
arcsec
x
−
sgn
(
x
)
ln
|
x
+
x
2
−
1
|
+
C
=
x
arcsec
x
+
sgn
(
x
)
ln
|
x
−
x
2
−
1
|
+
C
{\displaystyle \int \operatorname {arcsec} x{\mbox{d}}x=x\operatorname {arcsec} x-\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\operatorname {arcsec} x+\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C}
∫
arccsc
x
d
x
=
x
arccsc
x
+
sgn
(
x
)
ln
|
x
+
x
2
−
1
|
+
C
=
x
arccsc
x
−
sgn
(
x
)
ln
|
x
−
x
2
−
1
|
+
C
{\displaystyle \int \operatorname {arccsc} x{\mbox{d}}x=x\operatorname {arccsc} x+\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\operatorname {arccsc} x-\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C}
含有指數函數的積分
編輯
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}{\mbox{d}}x=e^{x}+C}
∫
α
x
d
x
=
α
x
ln
α
+
C
{\displaystyle \int \alpha ^{x}{\mbox{d}}x={\frac {\alpha ^{x}}{\ln \alpha }}+C}
∫
x
e
a
x
d
x
=
1
a
2
(
a
x
−
1
)
e
a
x
+
C
{\displaystyle \int xe^{ax}{\mbox{d}}x={\frac {1}{a^{2}}}(ax-1)e^{ax}+C}
∫
x
n
e
a
x
d
x
=
1
a
x
n
e
a
x
−
n
a
∫
x
n
−
1
e
a
x
d
x
{\displaystyle \int x^{n}e^{ax}{\mbox{d}}x={\frac {1}{a}}x^{n}e^{ax}-{\frac {n}{a}}\int x^{n-1}e^{ax}{\mbox{d}}x}
∫
e
a
x
sin
b
x
d
x
=
e
a
x
a
2
+
b
2
(
a
sin
b
x
−
b
cos
b
x
)
+
C
{\displaystyle \int e^{ax}\sin bx{\mbox{d}}x={\frac {e^{ax}}{a^{2}+b^{2}}}(a\sin bx-b\cos bx)+C}
∫
e
a
x
cos
b
x
d
x
=
e
a
x
a
2
+
b
2
(
a
cos
b
x
+
b
sin
b
x
)
+
C
{\displaystyle \int e^{ax}\cos bx{\mbox{d}}x={\frac {e^{ax}}{a^{2}+b^{2}}}(a\cos bx+b\sin bx)+C}
含有對數函數的積分
編輯
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
{\displaystyle \int \ln x{\mbox{d}}x=x\ln x-x+C}
∫
log
α
x
d
x
=
1
ln
α
(
x
ln
x
−
x
)
+
C
{\displaystyle \int \log _{\alpha }x{\mbox{d}}x={\frac {1}{\ln \alpha }}\left({x\ln x-x}\right)+C}
∫
x
n
ln
x
d
x
=
x
n
+
1
(
n
+
1
)
2
[
(
n
+
1
)
ln
x
−
1
]
+
C
{\displaystyle \int x^{n}\ln x{\mbox{d}}x={\frac {x^{n+1}}{(n+1)^{2}}}[(n+1)\ln x-1]+C}
∫
1
x
ln
x
d
x
=
ln
(
ln
x
)
+
C
{\displaystyle \int {\frac {1}{x\ln {x}}}{\mbox{d}}x=\ln {(\ln {x})}+C}
含有雙曲函數的積分
編輯
∫
sinh
x
d
x
=
cosh
x
+
C
{\displaystyle \int \sinh x{\mbox{d}}x=\cosh x+C}
∫
cosh
x
d
x
=
sinh
x
+
C
{\displaystyle \int \cosh x{\mbox{d}}x=\sinh x+C}
∫
tanh
x
d
x
=
ln
(
cosh
x
)
+
C
{\displaystyle \int \tanh x{\mbox{d}}x=\ln \left(\cosh x\right)+C}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
C
{\displaystyle \int \coth x{\mbox{d}}x=\ln \left|\sinh x\right|+C}
∫
sech
x
d
x
=
arcsin
(
tanh
x
)
+
C
=
arctan
(
sinh
x
)
+
C
{\displaystyle \int {\mbox{sech}}\ x{\mbox{d}}x=\arcsin \left(\tanh x\right)+C=\arctan \left(\sinh x\right)+C}
∫
csch
x
d
x
=
ln
|
tanh
x
2
|
+
C
{\displaystyle \int {\mbox{csch}}\ x{\mbox{d}}x=\ln \left|\tanh {x \over 2}\right|+C}
定積分
編輯
∫
−
∞
∞
e
−
α
x
2
d
x
=
π
α
{\displaystyle \int _{-\infty }^{\infty }e^{-\alpha x^{2}}{\mbox{d}}x={\sqrt {\frac {\pi }{\alpha }}}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
{
n
−
1
n
⋅
n
−
3
n
−
2
⋅
…
⋅
4
5
⋅
2
3
,
if
n
>
1
且
n
為 奇 數
n
−
1
n
⋅
n
−
3
n
−
2
⋅
…
⋅
3
4
⋅
1
2
⋅
π
2
,
if
n
>
0
且
n
為 偶 數
{\displaystyle \int _{0}^{\frac {\pi }{2}}{\mbox{sin}}^{n}x{\mbox{d}}x=\int _{0}^{\frac {\pi }{2}}{\mbox{cos}}^{n}x{\mbox{d}}x={\begin{cases}{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdot \ldots \cdot {\frac {4}{5}}\cdot {\frac {2}{3}},&{\mbox{if }}n>1{\mbox{ 且 }}n{\mbox{為 奇 數 }}\\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdot \ldots \cdot {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot {\frac {\pi }{2}},&{\mbox{if }}n>0{\mbox{ 且 }}n{\mbox{為 偶 數 }}\end{cases}}}
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