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包含 的积分
[edit | edit source]假设 ,对于 ,请参见下一节。
这里 ,其中 的正值需要被采用。
∫ x 4 d x s = x 3 s 4 + 3 8 a 2 x s + 3 8 a 4 ln | x + s a | {\displaystyle \int {\frac {x^{4}\;dx}{s}}={\frac {x^{3}s}{4}}+{\frac {3}{8}}a^{2}xs+{\frac {3}{8}}a^{4}\ln \left|{\frac {x+s}{a}}\right|}
∫ x 4 d x s 3 = x s 2 − a 2 x s + 3 2 a 2 ln | x + s a | {\displaystyle \int {\frac {x^{4}\;dx}{s^{3}}}={\frac {xs}{2}}-{\frac {a^{2}x}{s}}+{\frac {3}{2}}a^{2}\ln \left|{\frac {x+s}{a}}\right|}
∫ x 4 d x s 5 = − x s − 1 3 x 3 s 3 + ln | x + s a | {\displaystyle \int {\frac {x^{4}\;dx}{s^{5}}}=-{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}+\ln \left|{\frac {x+s}{a}}\right|}
∫ x 2 m d x s 2 n + 1 = ( − 1 ) n − m 1 a 2 ( n − m ) ∑ i = 0 n − m − 1 1 2 ( m + i ) + 1 ( n − m − 1 i ) x 2 ( m + i ) + 1 s 2 ( m + i ) + 1 ( n > m ≥ 0 ) {\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=(-1)^{n-m}{\frac {1}{a^{2(n-m)}}}\sum _{i=0}^{n-m-1}{\frac {1}{2(m+i)+1}}{n-m-1 \choose i}{\frac {x^{2(m+i)+1}}{s^{2(m+i)+1}}}\qquad {\mbox{(}}n>m\geq 0{\mbox{)}}}
∫ d x s 3 = − 1 a 2 x s {\displaystyle \int {\frac {dx}{s^{3}}}=-{\frac {1}{a^{2}}}{\frac {x}{s}}}
∫ d x s 5 = 1 a 4 [ x s − 1 3 x 3 s 3 ] {\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}
∫ d x s 7 = − 1 a 6 [ x s − 2 3 x 3 s 3 + 1 5 x 5 s 5 ] {\displaystyle \int {\frac {dx}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\frac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}
∫ d x s 9 = 1 a 8 [ x s − 3 3 x 3 s 3 + 3 5 x 5 s 5 − 1 7 x 7 s 7 ] {\displaystyle \int {\frac {dx}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\frac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}
∫ x 2 d x s 5 = − 1 a 2 x 3 3 s 3 {\displaystyle \int {\frac {x^{2}\;dx}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}}
∫ x 2 d x s 7 = 1 a 4 [ 1 3 x 3 s 3 − 1 5 x 5 s 5 ] {\displaystyle \int {\frac {x^{2}\;dx}{s^{7}}}={\frac {1}{a^{4}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}
∫ x 2 d x s 9 = − 1 a 6 [ 1 3 x 3 s 3 − 2 5 x 5 s 5 + 1 7 x 7 s 7 ] {\displaystyle \int {\frac {x^{2}\;dx}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}
∫ u d x = 1 2 ( x u + a 2 arcsin x a ) ( | x | ≤ | a | ) {\displaystyle \int u\;dx={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫ x u d x = − 1 3 u 3 ( | x | ≤ | a | ) {\displaystyle \int xu\;dx=-{\frac {1}{3}}u^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫ x 2 u d x = − x 4 u 3 + a 2 8 ( x u + a 2 arcsin x a ) ( | x | ≤ | a | ) {\displaystyle \int x^{2}u\;dx=-{\frac {x}{4}}u^{3}+{\frac {a^{2}}{8}}(xu+a^{2}\arcsin {\frac {x}{a}})\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
∫ d x u = arcsin x a ( | x | ≤ | a | ) {\displaystyle \int {\frac {dx}{u}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫ x 2 d x u = 1 2 ( − x u + a 2 arcsin x a ) ( | x | ≤ | a | ) {\displaystyle \int {\frac {x^{2}\;dx}{u}}={\frac {1}{2}}\left(-xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫ u d x = 1 2 ( x u − sgn x arcosh | x a | ) (for | x | ≥ | a | ) {\displaystyle \int u\;dx={\frac {1}{2}}\left(xu-\operatorname {sgn} x\,\operatorname {arcosh} \left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(for }}|x|\geq |a|{\mbox{)}}}
∫ x u d x = − u ( | x | ≤ | a | ) {\displaystyle \int {\frac {x}{u}}\;dx=-u\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
包含 R = a x 2 + b x + c {\displaystyle R={\sqrt {ax^{2}+bx+c}}} 的积分
[edit | edit source]假设 (ax2 + bx + c) 不能简化为 (px + q)2 的形式,其中 p 和 q 为任意常数。
∫ 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{\mbox{)}}}
∫ 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 ) , c > 0 {\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right),~c>0}
∫ d x x R = − 1 c arsinh ( b x + 2 c | x | 4 a c − b 2 ) , c < 0 {\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right),~c<0}
∫ d x x R = 1 − c arcsin ( b x + 2 c | x | b 2 − 4 a c ) , c < 0 , b 2 − 4 a c > 0 {\displaystyle \int {\frac {dx}{xR}}={\frac {1}{\sqrt {-c}}}\operatorname {arcsin} \left({\frac {bx+2c}{|x|{\sqrt {b^{2}-4ac}}}}\right),~c<0,b^{2}-4ac>0}
∫ d x x R = − 2 b x ( a x 2 + b x ) , c = 0 {\displaystyle \int {\frac {dx}{xR}}=-{\frac {2}{bx}}\left({\sqrt {ax^{2}+bx}}\right),~c=0}
(点击右侧的“显示”以查看证明,或点击“隐藏”以隐藏它。)
首先使用替换方法,令
,使得 u ≡ a x + b {\displaystyle u\equiv {\sqrt {ax+b}}} ,积分变为 R = x u {\displaystyle R={\sqrt {x}}u}
2 a ∫ d u ( u 2 − b ) 3 2 {\displaystyle 2{\sqrt {a}}\int {\frac {du}{(u^{2}-b)^{\frac {3}{2}}}}} 然后用
进行另一个替换,这样积分就变成了 v ≡ 1 u 2 − b {\displaystyle v\equiv {\frac {1}{\sqrt {u^{2}-b}}}}
− 2 a ∫ v d v 1 + b v 2 {\displaystyle -2{\sqrt {a}}\int {\frac {vdv}{\sqrt {1+bv^{2}}}}} 然后积分就具有
的形式,如上所示,可以很容易地计算出来。 ∫ x d x r = r {\displaystyle \int {\frac {x\;dx}{r}}=r} 也可以用不太明显的选项
,一步完成替换,得到 u ≡ a x + b a x {\displaystyle u\equiv {\sqrt {\frac {ax+b}{ax}}}}
∫ d x x R = − 2 a b ∫ d u = − 2 a b u {\displaystyle \int {\frac {dx}{xR}}=-{\frac {2{\sqrt {a}}}{b}}\int du=-{\frac {2{\sqrt {a}}}{b}}u}
涉及 S = a x + b {\displaystyle S={\sqrt {ax+b}}} 的积分
[edit | edit source]∫ S d x = 2 S 3 3 a {\displaystyle \int S{dx}={\frac {2S^{3}}{3a}}}
∫ d x S = 2 S a {\displaystyle \int {\frac {dx}{S}}={\frac {2S}{a}}}
∫ d x x S = { − 2 b a r c o t h ( S b ) (for b > 0 , a x > 0 ) − 2 b a r t a n h ( S b ) (for b > 0 , a x < 0 ) 2 − b arctan ( S − b ) (for b < 0 ) {\displaystyle \int {\frac {dx}{xS}}={\begin{cases}-{\frac {2}{\sqrt {b}}}\mathrm {arcoth} \left({\frac {S}{\sqrt {b}}}\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\-{\frac {2}{\sqrt {b}}}\mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)&{\mbox{(for }}b>0,\quad ax<0{\mbox{)}}\\{\frac {2}{\sqrt {-b}}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)&{\mbox{(for }}b<0{\mbox{)}}\\\end{cases}}}
∫ S x d x = { 2 ( S − b a r c o t h ( S b ) ) (for b > 0 , a x > 0 ) 2 ( S − b a r t a n h ( S b ) ) (for b > 0 , a x < 0 ) 2 ( S − − b arctan ( S − b ) ) (for b < 0 ) {\displaystyle \int {\frac {S}{x}}\,dx={\begin{cases}2\left(S-{\sqrt {b}}\,\mathrm {arcoth} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\2\left(S-{\sqrt {b}}\,\mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax<0{\mbox{)}}\\2\left(S-{\sqrt {-b}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)\right)&{\mbox{(for }}b<0{\mbox{)}}\\\end{cases}}}
∫ x n S d x = 2 a ( 2 n + 1 ) ( x n S − b n ∫ x n − 1 S d x ) {\displaystyle \int {\frac {x^{n}}{S}}dx={\frac {2}{a(2n+1)}}\left(x^{n}S-bn\int {\frac {x^{n-1}}{S}}dx\right)}
∫ x n S d x = 2 a ( 2 n + 3 ) ( x n S 3 − n b ∫ x n − 1 S d x ) {\displaystyle \int x^{n}Sdx={\frac {2}{a(2n+3)}}\left(x^{n}S^{3}-nb\int x^{n-1}Sdx\right)}
∫ 1 x n S d x = − 1 b ( n − 1 ) ( S x n − 1 + ( n − 3 2 ) a ∫ d x x n − 1 S ) {\displaystyle \int {\frac {1}{x^{n}S}}dx=-{\frac {1}{b(n-1)}}\left({\frac {S}{x^{n-1}}}+\left(n-{\frac {3}{2}}\right)a\int {\frac {dx}{x^{n-1}S}}\right)}