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:
"原始関数の一覧"
–
ニュース
·
書籍
·
スカラー
·
CiNii
·
J-STAGE
·
NDL
·
dlib.jp
·
ジャパンサーチ
·
TWL
(
2016年1月
)
本項は、
原始関数
の一覧
(げんしかんすうのいちらん)である。以下、積分定数は
C
{\displaystyle C}
とする。
目次
1
'"`UNIQ--postMath-00000002-QINU`"' を含む積分
2
'"`UNIQ--postMath-00000008-QINU`"' を含む積分
3
'"`UNIQ--postMath-00000011-QINU`"' を含む積分
4
'"`UNIQ--postMath-00000014-QINU`"' を含む積分
5
'"`UNIQ--postMath-00000016-QINU`"'を含む積分
6
'"`UNIQ--postMath-00000018-QINU`"' を含む積分
7
'"`UNIQ--postMath-00000021-QINU`"' を含む積分
8
'"`UNIQ--postMath-00000023-QINU`"' を含む積分
9
'"`UNIQ--postMath-0000002C-QINU`"' を含む積分
10
三角関数を含む積分
11
逆三角関数を含む積分
12
指数関数を含む積分
13
対数関数を含む積分
14
双曲線関数を含む積分
15
定積分
16
関連項目
a
x
+
b
{\displaystyle ax+b}
を含む積分
編集
∫
1
a
x
+
b
d
x
=
1
a
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {1}{ax+b}}\,dx={\frac {1}{a}}\ln \left|ax+b\right|+C}
∫
x
a
x
+
b
d
x
=
x
a
−
b
a
2
ln
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {x}{ax+b}}\,dx={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|+C}
∫
x
2
a
x
+
b
d
x
=
1
2
a
3
(
a
2
x
2
−
2
a
b
x
+
2
b
2
ln
|
a
x
+
b
|
)
+
C
{\displaystyle \int {\frac {x^{2}}{ax+b}}\,dx={\frac {1}{2a^{3}}}(a^{2}x^{2}-2abx+2b^{2}\ln \left|ax+b\right|)+C}
∫
1
x
(
a
x
+
b
)
d
x
=
−
1
b
ln
|
a
x
+
b
x
|
+
C
{\displaystyle \int {\frac {1}{x(ax+b)}}\,dx=-{\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)}}\,dx={\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|-{\frac {1}{bx}}+C}
a
+
b
x
{\displaystyle {\sqrt {a+bx}}}
を含む積分
編集
∫
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}}\,dx={\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}}\,dx={\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}}\,dx={\frac {2}{b(2n+3)}}x^{n}(a+bx)^{\frac {3}{2}}-{\frac {2na}{b(2n+3)}}\int x^{n-1}{\sqrt {a+bx}}dx}
∫
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}}\,dx=2{\sqrt {a+bx}}+a\int {\frac {1}{x{\sqrt {a+bx}}}}dx}
∫
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}}}\,dx={\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}}}dx,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}}}}\,dx={\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}
∫
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}}}}\,dx={\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}}dx,n\neq 1}
x
2
±
α
2
(
α
≠
0
)
{\displaystyle x^{2}\pm {\alpha }^{2}(\alpha \neq 0)}
を含む積分
編集
∫
1
x
2
+
α
2
d
x
=
1
α
arctan
x
α
+
C
{\displaystyle \int {\frac {1}{x^{2}+\alpha ^{2}}}\,dx={\frac {1}{\alpha }}\arctan {\frac {x}{\alpha }}+C}
∫
1
±
x
2
∓
α
2
d
x
=
1
2
α
ln
(
x
∓
α
±
x
+
α
)
+
C
{\displaystyle \int {\frac {1}{\pm x^{2}\mp \alpha ^{2}}}\,dx={\frac {1}{2\alpha }}\ln \left({\dfrac {x\mp \alpha }{\pm x+\alpha }}\right)+C}
a
x
2
+
b
{\displaystyle ax^{2}+b}
を含む積分
編集
∫
1
a
x
2
+
b
d
x
=
1
a
b
arctan
a
b
x
+
C
{\displaystyle \int {\frac {1}{ax^{2}+b}}\,dx={\frac {1}{\sqrt {ab}}}\arctan {\sqrt {\frac {a}{b}}}x+C}
a
x
2
+
b
x
+
c
(
a
≠
0
)
{\displaystyle ax^{2}+bx+c(a\neq 0)}
を含む積分
編集
∫
(
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)\,dx={\frac {ax^{3}}{3}}+{\frac {bx^{2}}{2}}+cx+C}
a
2
+
x
2
(
a
>
0
)
{\displaystyle {\sqrt {a^{2}+x^{2}}}\;(a>0)}
を含む積分
編集
∫
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}}}\,dx={\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}}}\,dx={\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}}\,dx={\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}}}\,dx=\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}}}}\,dx=\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}}}}\,dx={\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}}}}}\,dx={\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}}}}}\,dx=-{\frac {\sqrt {a^{2}+x^{2}}}{a^{2}x}}+C}
x
2
−
a
2
(
x
2
>
a
2
)
{\displaystyle {\sqrt {x^{2}-a^{2}}}\;(x^{2}>a^{2})}
を含む積分
編集
∫
1
x
2
−
a
2
d
x
=
ln
(
x
+
x
2
−
a
2
)
+
C
{\displaystyle \int {\frac {1}{\sqrt {x^{2}-a^{2}}}}\,dx=\ln \left(x+{\sqrt {x^{2}-a^{2}}}\right)+C}
a
2
−
x
2
(
a
2
>
x
2
)
{\displaystyle {\sqrt {a^{2}-x^{2}}}\;(a^{2}>x^{2})}
を含む積分
編集
∫
1
a
2
−
x
2
d
x
=
arcsin
x
a
+
C
=
−
arccos
x
a
+
C
{\displaystyle \int {\frac {1}{\sqrt {a^{2}-x^{2}}}}\,dx=\arcsin {\frac {x}{a}}+C=-\arccos {\frac {x}{a}}+C}
∫
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}}}\,dx={\frac {1}{2}}x{\sqrt {a^{2}-x^{2}}}+{\frac {a^{2}}{2}}\arcsin {\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}}}\,dx={\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}}\,dx={\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}}}\,dx=-{\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}}}}}\,dx=-{\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}}}}\,dx=-{\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}}}}}\,dx=-{\frac {\sqrt {a^{2}-x^{2}}}{a^{2}x}}+C}
R
=
|
a
|
x
2
+
b
x
+
c
(
a
≠
0
)
{\displaystyle R={\sqrt {|a|x^{2}+bx+c}}\;(a\neq 0)}
を含む積分
編集
∫
d
x
R
=
1
a
ln
(
2
a
R
+
2
a
x
+
b
)
(
for
a
>
0
)
{\displaystyle \int {\frac {dx}{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 {dx}{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 {dx}{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 {dx}{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 {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
arsinh
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}
三角関数を含む積分
編集
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos x\,dx=\sin x+C}
∫
−
sin
x
d
x
=
cos
x
+
C
{\displaystyle \int -\sin x\,dx=\cos x+C}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\tan x+C}
∫
−
csc
2
x
d
x
=
cot
x
+
C
{\displaystyle \int -\csc ^{2}x\,dx=\cot x+C}
∫
sec
x
tan
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec x\tan x\,dx=\sec x+C}
∫
−
csc
x
cot
x
d
x
=
csc
x
+
C
{\displaystyle \int -\csc x\cot x\,dx=\csc x+C}
∫
tan
x
d
x
=
−
ln
(
cos
x
)
+
C
{\displaystyle \int \tan x\,dx=-\ln(\cos x)+C}
∫
cot
x
d
x
=
ln
(
sin
x
)
+
C
{\displaystyle \int \cot x\,dx=\ln(\sin x)+C}
∫
sec
x
d
x
=
ln
(
sec
x
+
tan
x
)
+
C
=
gd
−
1
x
+
C
gd
−
1
x
{\displaystyle \int \sec x\,dx=\ln(\sec x+\tan x)+C=\operatorname {gd} ^{-1}x+C\quad \operatorname {gd} ^{-1}x}
:
グーデルマン関数
の
逆関数
∫
csc
x
d
x
=
−
ln
(
csc
x
+
cot
x
)
+
C
=
ln
(
tan
x
−
sin
x
sin
x
tan
x
)
+
C
{\displaystyle \int \csc x\,dx=-\ln(\csc x+\cot x)+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\,dx=-{\frac {1}{n}}\sin ^{n-1}x\cos x+{\frac {n-1}{n}}\int \sin ^{n-2}x\,dx+C\quad \forall n\geq 2}
∫
sin
2
x
d
x
=
x
2
−
sin
2
x
4
+
C
{\displaystyle \int \sin ^{2}x\,dx={\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\,dx={\frac {1}{n}}\cos ^{n-1}x\sin x+{\frac {n-1}{n}}\int \cos ^{n-2}x\,dx+C\quad \forall n\geq 2}
∫
cos
2
x
d
x
=
x
2
+
sin
2
x
4
+
C
{\displaystyle \int \cos ^{2}x\,dx={\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\,dx={\frac {1}{n-1}}\tan ^{n-1}x-\int \tan ^{n-2}x\,dx+C\quad \forall n\geq 2}
∫
tan
2
x
d
x
=
tan
x
−
x
+
C
{\displaystyle \int \tan ^{2}x\,dx=\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\,dx={\frac {1}{n-1}}\cot ^{n-1}x-\int \cot ^{n-2}x\,dx+C\quad \forall n\geq 2}
∫
cot
2
x
d
x
=
−
cot
x
−
x
+
C
{\displaystyle \int \cot ^{2}x\,dx=-\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\,dx={\frac {1}{n-1}}\sec ^{n-2}x\tan x+{\frac {n-2}{n-1}}\int \sec ^{n-2}x\,dx+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\,dx=-{\frac {1}{n-1}}\csc ^{n-2}x\cot x+{\frac {n-2}{n-1}}\int \csc ^{n-2}x\,dx+C\quad \forall n\geq 2}
逆三角関数を含む積分
編集
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
C
{\displaystyle \int \arcsin x\,dx=x\arcsin x+{\sqrt {1-x^{2}}}+C}
∫
arccos
x
d
x
=
x
arccos
x
−
1
−
x
2
+
C
{\displaystyle \int \arccos x\,dx=x\arccos x-{\sqrt {1-x^{2}}}+C}
∫
arctan
x
d
x
=
x
arctan
x
−
ln
1
+
x
2
+
C
{\displaystyle \int \arctan x\,dx=x\arctan x-\ln {\sqrt {1+x^{2}}}+C}
∫
arccot
x
d
x
=
x
arccot
x
+
ln
1
+
x
2
+
C
{\displaystyle \int \operatorname {arccot} x\,dx=x\operatorname {arccot} x+\ln {\sqrt {1+x^{2}}}+C}
∫
arcsec
x
d
x
=
x
arcsec
x
−
ln
(
x
−
x
2
−
1
)
+
C
{\displaystyle \int \operatorname {arcsec} x\,dx=x\operatorname {arcsec} x-\ln(x-{\sqrt {x^{2}-1}})+C}
∫
arccsc
x
d
x
=
x
arccsc
x
+
ln
(
x
+
x
2
−
1
)
+
C
{\displaystyle \int \operatorname {arccsc} x\,dx=x\operatorname {arccsc} x+\ln(x+{\sqrt {x^{2}-1}})+C}
指数関数を含む積分
編集
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}\,dx=e^{x}+C}
∫
α
x
d
x
=
α
x
ln
α
+
C
{\displaystyle \int \alpha ^{x}\,dx={\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}\,dx={\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}\,dx={\frac {1}{a}}x^{n}e^{ax}-{\frac {n}{a}}\int x^{n-1}e^{ax}\,dx}
∫
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\,dx={\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\,dx={\frac {e^{ax}}{a^{2}+b^{2}}}(a\cos bx+b\sin bx)+C}
対数関数を含む積分
編集
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
{\displaystyle \int \ln x\,dx=x\ln x-x+C}
∫
log
α
x
d
x
=
1
ln
α
(
x
ln
x
−
x
)
+
C
{\displaystyle \int \log _{\alpha }x\,dx={\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\,dx={\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}}}\,dx=\ln {(\ln {x})}+C}
双曲線関数を含む積分
編集
∫
sinh
x
d
x
=
cosh
x
+
C
{\displaystyle \int \sinh x\,dx=\cosh x+C}
∫
cosh
x
d
x
=
sinh
x
+
C
{\displaystyle \int \cosh x\,dx=\sinh x+C}
∫
tanh
x
d
x
=
ln
(
cosh
x
)
+
C
{\displaystyle \int \tanh x\,dx=\ln \left(\cosh x\right)+C}
∫
coth
x
d
x
=
ln
(
sinh
x
)
+
C
{\displaystyle \int \coth x\,dx=\ln \left(\sinh x\right)+C}
∫
sech
x
d
x
=
arcsin
(
tanh
x
)
+
C
=
arctan
(
sinh
x
)
+
C
=
gd
x
+
C
gd
x
{\displaystyle \int {\mbox{sech}}\ x\,dx=\arcsin \left(\tanh x\right)+C=\arctan \left(\sinh x\right)+C=\operatorname {gd} x+C\quad \operatorname {gd} x}
:
グーデルマン関数
∫
csch
x
d
x
=
ln
(
tanh
x
2
)
+
C
{\displaystyle \int {\mbox{csch}}\ x\,dx=\ln \left(\tanh {x \over 2}\right)+C}
定積分
編集
∫
−
∞
∞
e
−
α
x
2
d
x
=
π
α
{\displaystyle \int _{-\infty }^{\infty }e^{-\alpha x^{2}}\,dx={\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
and
n
is odd
n
−
1
n
⋅
n
−
3
n
−
2
⋅
⋯
⋅
3
4
⋅
1
2
⋅
π
2
,
if
n
>
0
and
n
is even
{\displaystyle \int _{0}^{\frac {\pi }{2}}{\mbox{sin}}^{n}x\,dx=\int _{0}^{\frac {\pi }{2}}{\mbox{cos}}^{n}x\,dx={\begin{cases}{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdot \cdots \cdot {\frac {4}{5}}\cdot {\frac {2}{3}},&{\mbox{if }}n>1{\mbox{ and }}n{\mbox{ is odd}}\\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdot \cdots \cdot {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot {\frac {\pi }{2}},&{\mbox{if }}n>0{\mbox{ and }}n{\mbox{ is even}}\end{cases}}}
関連項目
編集
プロジェクト 数学
ポータル 数学
微分積分学の基礎定理
解析学