有理関数の原始関数の一覧

本項は、有理関数の原始関数の一覧である。さらに完全な原始関数の一覧は、原始関数の一覧を参照のこと。

∫ ( a x + b ) n d x = ( a x + b ) n + 1 a ( n + 1 ) + C (for  n ≠ − 1 ) {\displaystyle \int (ax+b)^{n}dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫ c a x + b d x = c a ln ⁡ | a x + b | + C {\displaystyle \int {\frac {c}{ax+b}}dx={\frac {c}{a}}\ln \left|ax+b\right|+C}
∫ x ( a x + b ) n d x = a ( n + 1 ) x − b a 2 ( n + 1 ) ( n + 2 ) ( a x + b ) n + 1 + C (for  n ∉ { − 1 , − 2 } ) {\displaystyle \int x(ax+b)^{n}dx={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}+C\qquad {\mbox{(for }}n\not \in \{-1,-2\}{\mbox{)}}}


∫ 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 ( a x + b ) 2 d x = b a 2 ( a x + b ) + 1 a 2 ln ⁡ | a x + b | + C {\displaystyle \int {\frac {x}{(ax+b)^{2}}}dx={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|+C}
∫ x ( a x + b ) n d x = a ( 1 − n ) x − b a 2 ( n − 1 ) ( n − 2 ) ( a x + b ) n − 1 + C (for  n ∉ { 1 , 2 } ) {\displaystyle \int {\frac {x}{(ax+b)^{n}}}dx={\frac {a(1-n)x-b}{a^{2}(n-1)(n-2)(ax+b)^{n-1}}}+C\qquad {\mbox{(for }}n\not \in \{1,2\}{\mbox{)}}}


∫ f ′ ( x ) f ( x ) d x = ln ⁡ | f ( x ) | + C {\displaystyle \int {\frac {f'(x)}{f(x)}}dx=\ln \left|f(x)\right|+C}


∫ x 2 a x + b d x = b 2 ln ⁡ ( | a x + b | ) a 3 + a x 2 − 2 b x 2 a 2 + C {\displaystyle \int {\frac {x^{2}}{ax+b}}dx={\frac {b^{2}\ln(\left|ax+b\right|)}{a^{3}}}+{\frac {ax^{2}-2bx}{2a^{2}}}+C}
∫ x 2 ( a x + b ) 2 d x = 1 a 3 ( a x − 2 b ln ⁡ | a x + b | − b 2 a x + b ) + C {\displaystyle \int {\frac {x^{2}}{(ax+b)^{2}}}dx={\frac {1}{a^{3}}}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)+C}
∫ x 2 ( a x + b ) 3 d x = 1 a 3 ( ln ⁡ | a x + b | + 2 b a x + b − b 2 2 ( a x + b ) 2 ) + C {\displaystyle \int {\frac {x^{2}}{(ax+b)^{3}}}dx={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)+C}
∫ x 2 ( a x + b ) n d x = 1 a 3 ( − ( a x + b ) 3 − n ( n − 3 ) + 2 b ( a x + b ) 2 − n ( n − 2 ) − b 2 ( a x + b ) 1 − n ( n − 1 ) ) + C (for  n ∉ { 1 , 2 , 3 } ) {\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx={\frac {1}{a^{3}}}\left(-{\frac {(ax+b)^{3-n}}{(n-3)}}+{\frac {2b(ax+b)^{2-n}}{(n-2)}}-{\frac {b^{2}(ax+b)^{1-n}}{(n-1)}}\right)+C\qquad {\mbox{(for }}n\not \in \{1,2,3\}{\mbox{)}}}


∫ 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 = − 1 b x + a b 2 ln ⁡ | a x + b x | + C {\displaystyle \int {\frac {1}{x^{2}(ax+b)}}dx=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C}
∫ 1 x 2 ( a x + b ) 2 d x = − a ( 1 b 2 ( a x + b ) + 1 a b 2 x − 2 b 3 ln ⁡ | a x + b x | ) + C {\displaystyle \int {\frac {1}{x^{2}(ax+b)^{2}}}dx=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)+C}
∫ 1 x 2 + a 2 d x = 1 a arctan ⁡ x a + C {\displaystyle \int {\frac {1}{x^{2}+a^{2}}}dx={\frac {1}{a}}\arctan {\frac {x}{a}}\,\!+C}
∫ 1 x 2 − a 2 d x = { − 1 a a r c t a n h x a = 1 2 a ln ⁡ a − x a + x + C (for  | x | < | a | ) − 1 a a r c c o t h x a = 1 2 a ln ⁡ x − a x + a + C (for  | x | > | a | ) {\displaystyle \int {\frac {1}{x^{2}-a^{2}}}dx={\begin{cases}-{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C&{\mbox{(for }}|x|<|a|{\mbox{)}}\\-{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C&{\mbox{(for }}|x|>|a|{\mbox{)}}\end{cases}}}


For a ≠ 0 : {\displaystyle a\neq 0:}

∫ 1 a x 2 + b x + c d x = { 2 4 a c − b 2 arctan ⁡ 2 a x + b 4 a c − b 2 + C (for  4 a c − b 2 > 0 ) − 2 b 2 − 4 a c a r c t a n h 2 a x + b b 2 − 4 a c + C = 1 b 2 − 4 a c ln ⁡ | 2 a x + b − b 2 − 4 a c 2 a x + b + b 2 − 4 a c | + C (for  4 a c − b 2 < 0 ) − 2 2 a x + b + C (for  4 a c − b 2 = 0 ) {\displaystyle \int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}{\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\mbox{(for }}4ac-b^{2}>0{\mbox{)}}\\-{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C&{\mbox{(for }}4ac-b^{2}<0{\mbox{)}}\\-{\frac {2}{2ax+b}}+C&{\mbox{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}


∫ x a x 2 + b x + c d x = 1 2 a ln ⁡ | a x 2 + b x + c | − b 2 a ∫ d x a x 2 + b x + c + C {\displaystyle \int {\frac {x}{ax^{2}+bx+c}}dx={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {dx}{ax^{2}+bx+c}}+C}


∫ m x + n a x 2 + b x + c d x = { m 2 a ln ⁡ | a x 2 + b x + c | + 2 a n − b m a 4 a c − b 2 arctan ⁡ 2 a x + b 4 a c − b 2 + C (for  4 a c − b 2 > 0 ) m 2 a ln ⁡ | a x 2 + b x + c | − 2 a n − b m a b 2 − 4 a c a r c t a n h 2 a x + b b 2 − 4 a c + C (for  4 a c − b 2 < 0 ) m 2 a ln ⁡ | a x 2 + b x + c | − 2 a n − b m a ( 2 a x + b ) + C (for  4 a c − b 2 = 0 ) {\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\begin{cases}{\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\mbox{(for }}4ac-b^{2}>0{\mbox{)}}\\{\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C&{\mbox{(for }}4ac-b^{2}<0{\mbox{)}}\\{\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C&{\mbox{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}}


∫ 1 ( a x 2 + b x + c ) n d x = 2 a x + b ( n − 1 ) ( 4 a c − b 2 ) ( a x 2 + b x + c ) n − 1 + ( 2 n − 3 ) 2 a ( n − 1 ) ( 4 a c − b 2 ) ∫ 1 ( a x 2 + b x + c ) n − 1 d x + C {\displaystyle \int {\frac {1}{(ax^{2}+bx+c)^{n}}}dx={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {(2n-3)2a}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx+C}
∫ x ( a x 2 + b x + c ) n d x = − b x + 2 c ( n − 1 ) ( 4 a c − b 2 ) ( a x 2 + b x + c ) n − 1 − b ( 2 n − 3 ) ( n − 1 ) ( 4 a c − b 2 ) ∫ 1 ( a x 2 + b x + c ) n − 1 d x + C {\displaystyle \int {\frac {x}{(ax^{2}+bx+c)^{n}}}dx=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {b(2n-3)}{(n-1)(4ac-b^{2})}}\int {\frac {1}{(ax^{2}+bx+c)^{n-1}}}dx+C}
∫ 1 x ( a x 2 + b x + c ) d x = 1 2 c ln ⁡ | x 2 a x 2 + b x + c | − b 2 c ∫ 1 a x 2 + b x + c d x + C {\displaystyle \int {\frac {1}{x(ax^{2}+bx+c)}}dx={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {1}{ax^{2}+bx+c}}dx+C}


∫ d x x 2 n + 1 = ∑ k = 1 2 n − 1 { 1 2 n − 1 [ sin ⁡ ( ( 2 k − 1 ) π 2 n ) arctan ⁡ [ ( x − cos ⁡ ( ( 2 k − 1 ) π 2 n ) ) csc ⁡ ( ( 2 k − 1 ) π 2 n ) ] ] − 1 2 n [ cos ⁡ ( ( 2 k − 1 ) π 2 n ) ln ⁡ | x 2 − 2 x cos ⁡ ( ( 2 k − 1 ) π 2 n ) + 1 | ] } {\displaystyle \int {\frac {dx}{x^{2^{n}}+1}}=\sum _{k=1}^{2^{n-1}}\left\{{\frac {1}{2^{n-1}}}\left[\sin \left({\frac {(2k-1)\pi }{2^{n}}}\right)\arctan \left[\left(x-\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right)\csc \left({\frac {(2k-1)\pi }{2^{n}}}\right)\right]\right]-{\frac {1}{2^{n}}}\left[\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)\ln \left|x^{2}-2x\cos \left({\frac {(2k-1)\pi }{2^{n}}}\right)+1\right|\right]\right\}}

全ての有理関数は上記の公式を用いるか、または部分分数分解を行い、以下の形に変形することで積分を行うことができる。

e x + f ( a x 2 + b x + c ) n {\displaystyle {\frac {ex+f}{\left(ax^{2}+bx+c\right)^{n}}}} .