有理函数积分表此条目没有列出任何参考或来源。 (2017年12月26日)维基百科所有的内容都应该可供查证。请协助补充可靠来源以改善这篇条目。无法查证的内容可能会因为异议提出而移除。以下是部分有理函数的积分表。 ∫ ( a x + b ) n d x = ( a x + b ) n + 1 a ( n + 1 ) + C ( n ≠ − 1 ) {\displaystyle \int (ax+b)^{n}dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad (n\neq -1)} ∫ 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 ) n d x = a ( n + 1 ) x − b a 2 ( n + 1 ) ( n + 2 ) ( a x + b ) n + 1 + C ( 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 (n\not \in \{1,2\})} ∫ 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 ( 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 (n\not \in \{1,2\})} ∫ x 2 a x + b d x = 1 a 3 [ ( a x + b ) 2 2 − 2 b ( a x + b ) + b 2 ln | a x + b | ] + C {\displaystyle \int {\frac {x^{2}}{ax+b}}dx={\frac {1}{a^{3}}}\left[{\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right]+C} ∫ x 2 ( a x + b ) 2 d x = 1 a 3 ( a x + b − 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+b-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 [ − 1 ( n − 3 ) ( a x + b ) n − 3 + 2 b ( n − 2 ) ( a x + b ) n − 2 − b 2 ( n − 1 ) ( a x + b ) n − 1 ] + C {\displaystyle \int {\frac {x^{2}}{(ax+b)^{n}}}dx={\frac {1}{a^{3}}}\left[-{\frac {1}{(n-3)(ax+b)^{n-3}}}+{\frac {2b}{(n-2)(ax+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right]+C\qquad } ( n ∉ { 1 , 2 , 3 } ) {\displaystyle (n\not \in \{1,2,3\})} ∫ d x x ( a x + b ) = − 1 b ln | a x + b x | + C {\displaystyle \int {\frac {dx}{x(ax+b)}}=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C} ∫ d x x 2 ( a x + b ) = − 1 b x + a b 2 ln | a x + b x | + C {\displaystyle \int {\frac {dx}{x^{2}(ax+b)}}=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C} ∫ d x x 2 ( a x + b ) 2 = − a [ 1 b 2 ( a x + b ) + 1 a b 2 x − 2 b 3 ln | a x + b x | ] + C {\displaystyle \int {\frac {dx}{x^{2}(ax+b)^{2}}}=-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} ∫ d x x 2 + a 2 = 1 a arctan x a + C {\displaystyle \int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C} ∫ d x x 2 − a 2 = − 1 a a r c t a n h x a = 1 2 a ln a − x a + x + C ( | x | < | a | ) {\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arctanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C\qquad {\mbox{(}}|x|<|a|{\mbox{)}}\,\!} ∫ d x x 2 − a 2 = − 1 a a r c c o t h x a = 1 2 a ln x − a x + a + C ( | x | > | a | ) {\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arccoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C\qquad {\mbox{(}}|x|>|a|{\mbox{)}}\,\!} ∫ d x a x 2 + b x + c = 2 4 a c − b 2 arctan 2 a x + b 4 a c − b 2 + C , ( 4 a c − b 2 > 0 ) {\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C,(4ac-b^{2}>0)} ∫ d x a x 2 + b x + c = 2 b 2 − 4 a c a r 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 ( 4 a c − b 2 < 0 ) {\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}={\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {artanh} {\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\qquad {\mbox{(}}4ac-b^{2}<0{\mbox{)}}} ∫ d x a x 2 + b x + c = − 2 2 a x + b + C ( 4 a c − b 2 = 0 ) {\displaystyle \int {\frac {dx}{ax^{2}+bx+c}}=-{\frac {2}{2ax+b}}+C\qquad {\mbox{(}}4ac-b^{2}=0{\mbox{)}}} ∫ 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 ( 4 a c − b 2 > 0 ) {\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\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\qquad {\mbox{(}}4ac-b^{2}>0{\mbox{)}}} ∫ 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 b 2 − 4 a c a r t a n h 2 a x + b b 2 − 4 a c + C ( 4 a c − b 2 < 0 ) {\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|+{\frac {2an-bm}{a{\sqrt {b^{2}-4ac}}}}\,\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C\qquad {\mbox{(}}4ac-b^{2}<0{\mbox{)}}} ∫ 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 ( 2 a x + b ) + C ( 4 a c − b 2 = 0 ) {\displaystyle \int {\frac {mx+n}{ax^{2}+bx+c}}dx={\frac {m}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {2an-bm}{a(2ax+b)}}+C\qquad {\mbox{(}}4ac-b^{2}=0{\mbox{)}}} ∫ d x ( a x 2 + b x + c ) n = 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 ) ∫ d x ( a x 2 + b x + c ) n − 1 + C {\displaystyle \int {\frac {dx}{(ax^{2}+bx+c)^{n}}}={\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 {dx}{(ax^{2}+bx+c)^{n-1}}}+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 ) ∫ d x ( a x 2 + b x + c ) n − 1 + 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 {dx}{(ax^{2}+bx+c)^{n-1}}}+C\,\!} 对于任意的有理函数，我们都能通过部分分式(partial fraction)把该函数分拆为数个函数的总和，其中每个函数符合以下的形式： p x + q ( a x 2 + b x + c ) n {\displaystyle {\frac {px+q}{\left(ax^{2}+bx+c\right)^{n}}}} 。我们继而能把每一个该种形式的函数作积分运算： ∫ d x x ( a x 2 + b x + c ) = 1 2 c ln | x 2 a x 2 + b x + c | − b 2 c ∫ d x a x 2 + b x + c + C {\displaystyle \int {\frac {dx}{x(ax^{2}+bx+c)}}={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {dx}{ax^{2}+bx+c}}+C}
![\int\frac{x^2}{ax + b}dx = \frac{1}{a^3}\left[\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right] +C](/media/math_img/838/7fcd8cdc3fcc49bbb960ac38577fced7901a8c26.svg)
![\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](/media/math_img/838/629e71abe917e170f3031c9a5d6993bb16750dd0.svg)
![\int\frac{x^2}{(ax + b)^n}dx = \frac{1}{a^3}\left[-\frac{1}{(n- 3)(ax + b)^{n-3}} + \frac{2b}{(n-2)(ax + b)^{n-2}} - \frac{b^2}{(n - 1)(ax + b)^{n-1}}\right] +C \qquad](/media/math_img/838/30c63646a64dffc23fc3f27f3806c95c20c53500.svg)
![\int\frac{dx}{x^2(ax+b)^2} = -a\left[\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right] +C](/media/math_img/838/ab677379195030a5800e547aa90b86600ba4ca57.svg)
。我们继而能把每一个该种形式的函数作积分运算：
