Home » Posts 有理関数の原始関数の一覧 March 29, 2025 · 3 min · 591 words · Katie Triggs 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}}}} .
