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。 目录 1 含有'"`UNIQ--postMath-00000001-QINU`"'的积分 2 含有'"`UNIQ--postMath-00000008-QINU`"'的积分 3 含有'"`UNIQ--postMath-00000012-QINU`"'的积分 4 含有 '"`UNIQ--postMath-00000015-QINU`"'的积分 5 含有 '"`UNIQ--postMath-00000017-QINU`"'的积分 6 含有 '"`UNIQ--postMath-00000019-QINU`"'的积分 7 含有'"`UNIQ--postMath-00000022-QINU`"'的积分 8 含有'"`UNIQ--postMath-00000024-QINU`"'的积分 9 含有'"`UNIQ--postMath-0000002D-QINU`"'的积分 10 含有三角函数的积分 11 含有反三角函数的积分 12 含有指数函数的积分 13 含有对数函数的积分 14 含有双曲函数的积分 15 定积分 由于列表比较长，积分表被分为以下几个部分： 有理函数积分表 无理函数积分表 指数函数积分表 对数函数积分表 高斯函数积分表 三角函数积分表 反三角函数积分表 双曲函数积分表 反双曲函数积分表含有 a x + b {\displaystyle ax+b} 的积分 ∫ ( a x + b ) n d x = ( a x + b ) n + 1 a ( n + 1 ) + C {\displaystyle \int \ (ax+b)^{n}{\mbox{d}}x={\frac {(ax+b)^{n+1}}{a(n+1)}}+C} ∫ 1 a x + b d x = 1 a ln | a x + b | + C {\displaystyle \int {\frac {1}{ax+b}}{\mbox{d}}x={\frac {1}{a}}\ln \left|ax+b\right|+C} ∫ x a x + b d x = 1 a 2 ( a x + b − b ln | a x + b | ) + C {\displaystyle \int {\frac {x}{ax+b}}{\mbox{d}}x={\frac {1}{a^{2}}}(ax+b-b\ln \left|ax+b\right|)+C} ∫ x 2 a x + b d x = 1 2 a 3 [ ( a x + b ) 2 − 4 b ( a x + b ) + 2 b 2 ln | a x + b | ] + C {\displaystyle \int {\frac {x^{2}}{ax+b}}{\mbox{d}}x={\frac {1}{2a^{3}}}\left[(ax+b)^{2}-4b(ax+b)+2b^{2}\ln \left|ax+b\right|\right]+C} ∫ 1 x ( a x + b ) d x = − 1 b ln | a x + b x | + C {\displaystyle \int {\frac {1}{x(ax+b)}}{\mbox{d}}x=-{\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)}}{\mbox{d}}x={\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|-{\frac {1}{bx}}+C} 含有 a + b x {\displaystyle {\sqrt {a+bx}}} 的积分 ∫ 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}}{\mbox{d}}x={\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}}{\mbox{d}}x={\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}}{\mbox{d}}x={\frac {2}{b(2n+3)}}x^{n}(a+bx)^{\frac {3}{2}}-{\frac {2na}{b(2n+3)}}\int x^{n-1}{\sqrt {a+bx}}{\mbox{d}}x} ∫ 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}}{\mbox{d}}x=2{\sqrt {a+bx}}+a\int {\frac {1}{x{\sqrt {a+bx}}}}{\mbox{d}}x} ∫ 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}}}{\mbox{d}}x={\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}}}{\mbox{d}}x,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}}}}{\mbox{d}}x={\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} ∫ x a + b x d x = 2 ( a + b x ) 3 2 3 b 2 − ( 2 a ) a + b x b 2 {\displaystyle \int {\frac {x}{\sqrt {a+bx}}}{\mbox{d}}x={\frac {2(a+bx)^{\frac {3}{2}}}{3b^{2}}}-{\frac {(2a){\sqrt {a+bx}}}{b^{2}}}} ∫ 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}}}}{\mbox{d}}x={\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}}{\mbox{d}}x,n\neq 1} 含有 x 2 ± α 2 {\displaystyle x^{2}\pm \alpha ^{2}} 的积分 ∫ 1 x 2 + α 2 d x = arctan x α α + C {\displaystyle \int {\frac {1}{x^{2}+\alpha ^{2}}}{\mbox{d}}x={\frac {\arctan {\dfrac {x}{\alpha }}}{\alpha }}+C} ∫ 1 ± x 2 ∓ α 2 d x = ln ( x ∓ α ± x + α ) 2 α + C {\displaystyle \int {\frac {1}{\pm x^{2}\mp \alpha ^{2}}}{\mbox{d}}x={\frac {\ln \left({\dfrac {x\mp \alpha }{\pm x+\alpha }}\right)}{2\alpha }}+C} 含有 a x 2 + b {\displaystyle {ax^{2}+b}} 的积分 ∫ 1 a x 2 + b d x = 1 a b arctan a x b + C {\displaystyle \int {\frac {1}{ax^{2}+b}}{\mbox{d}}x={\frac {1}{\sqrt {ab}}}\arctan {\frac {{\sqrt {a}}x}{\sqrt {b}}}+C} 含有 a x 2 + b x + c {\displaystyle ax^{2}+bx+c} 的积分 ∫ 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{\mbox{d}}x={\frac {ax^{3}}{3}}+{\frac {bx^{2}}{2}}+cx+C} 含有 a 2 + x 2 ( a > 0 ) {\displaystyle {\sqrt {a^{2}+x^{2}}}\qquad (a>0)} 的积分 ∫ 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}}}{\mbox{d}}x={\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}}}{\mbox{d}}x={\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}}{\mbox{d}}x={\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}}}{\mbox{d}}x=\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}}}}{\mbox{d}}x=\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}}}}{\mbox{d}}x={\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}}}}}{\mbox{d}}x={\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}}}}}{\mbox{d}}x=-{\frac {\sqrt {a^{2}+x^{2}}}{a^{2}x}}+C} 含有 x 2 − a 2 ( x 2 > a 2 ) {\displaystyle {\sqrt {x^{2}-a^{2}}}\qquad {(x^{2}>a^{2})}} 的积分 ∫ 1 x 2 − a 2 d x = l n | x + x 2 − a 2 | + C {\displaystyle \int {\frac {1}{\sqrt {x^{2}-a^{2}}}}{\mbox{d}}x=ln|x+{\sqrt {x^{2}-a^{2}}}|+C} 含有 a 2 − x 2 ( a 2 > x 2 ) {\displaystyle {\sqrt {a^{2}-x^{2}}}\qquad (a^{2}>x^{2})} 的积分 ∫ 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}}}{\mbox{d}}x={\frac {1}{2}}x{\sqrt {a^{2}-x^{2}}}+{\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+C} ∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C {\displaystyle \int {\frac {1}{\sqrt {a^{2}-x^{2}}}}{\mbox{d}}x=\arcsin {\frac {x}{a}}+C=-\arccos {\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}}}{\mbox{d}}x={\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}}{\mbox{d}}x={\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}}}{\mbox{d}}x=-{\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}}}}}{\mbox{d}}x=-{\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}}}}{\mbox{d}}x=-{\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}}}}}{\mbox{d}}x=-{\frac {\sqrt {a^{2}-x^{2}}}{a^{2}x}}+C} 含有 R = | a | x 2 + b x + c ( a ≠ 0 ) {\displaystyle R={\sqrt {|a|x^{2}+bx+c}}\qquad (a\neq 0)} 的积分 ∫ d x R = 1 a ln ( 2 a R + 2 a x + b ) ( for a > 0 ) {\displaystyle \int {\frac {{\mbox{d}}x}{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 {{\mbox{d}}x}{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 {{\mbox{d}}x}{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 {{\mbox{d}}x}{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 {{\mbox{d}}x}{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 {{\mbox{d}}x}{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 {{\mbox{d}}x}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left[{\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {{\mbox{d}}x}{R^{2n-1}}}\right]} ∫ x R d x = R a − b 2 a ∫ d x R {\displaystyle \int {\frac {x}{R}}\;{\mbox{d}}x={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {{\mbox{d}}x}{R}}} ∫ x R 3 d x = − 2 b x + 4 c ( 4 a c − b 2 ) R {\displaystyle \int {\frac {x}{R^{3}}}\;{\mbox{d}}x=-{\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}}}\;{\mbox{d}}x=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {{\mbox{d}}x}{R^{2n+1}}}} ∫ d x x R = − 1 c ln ( 2 c R + b x + 2 c x ) {\displaystyle \int {\frac {{\mbox{d}}x}{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 {{\mbox{d}}x}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)} 含有三角函数的积分 ∫ cos x d x = sin x + C {\displaystyle \int \cos x{\mbox{d}}x=\sin x+C} ∫ sin x d x = − cos x + C {\displaystyle \int \sin x{\mbox{d}}x=-\cos x+C} ∫ sec 2 x d x = tan x + C {\displaystyle \int \sec ^{2}x{\mbox{d}}x=\tan x+C} ∫ csc 2 x d x = − cot x + C {\displaystyle \int \csc ^{2}x{\mbox{d}}x=-\cot x+C} ∫ sec x tan x d x = sec x + C {\displaystyle \int \sec x\tan x{\mbox{d}}x=\sec x+C} ∫ csc x cot x d x = − csc x + C {\displaystyle \int \csc x\cot x{\mbox{d}}x=-\csc x+C} ∫ tan x d x = − ln | cos x | + C = ln | sec x | + C {\displaystyle \int \tan x{\mbox{d}}x=-\ln {\left|\cos {x}\right|}+C=\ln {\left|\sec x\right|}+C} ∫ cot x d x = ln | sin x | + C {\displaystyle \int \cot x{\mbox{d}}x=\ln {\left|\sin x\right|}+C} ∫ sec x d x = ln | sec x + tan x | + C {\displaystyle \int \sec x{\mbox{d}}x=\ln {\left|\sec x+\tan x\right|}+C} ∫ csc x d x = ln | csc x − cot x | + C = ln | tan x − sin x sin x tan x | + C {\displaystyle \int \csc x{\mbox{d}}x=\ln {\left|\csc x-\cot x\right|}+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{\mbox{d}}x=-{\frac {1}{n}}\sin ^{n-1}x\cos x+{\frac {n-1}{n}}\int \sin ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2} ∫ sin 2 x d x = x 2 − sin 2 x 4 + C {\displaystyle \int \sin ^{2}x{\mbox{d}}x={\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{\mbox{d}}x={\frac {1}{n}}\cos ^{n-1}x\sin x+{\frac {n-1}{n}}\int \cos ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2} ∫ cos 2 x d x = x 2 + sin 2 x 4 + C {\displaystyle \int \cos ^{2}x{\mbox{d}}x={\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{\mbox{d}}x={\frac {1}{n-1}}\tan ^{n-1}x-\int \tan ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2} ∫ tan 2 x d x = tan x − x + C {\displaystyle \int \tan ^{2}x{\mbox{d}}x=\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{\mbox{d}}x=-{\frac {1}{n-1}}\cot ^{n-1}x-\int \cot ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2} ∫ cot 2 x d x = − cot x − x + C {\displaystyle \int \cot ^{2}x{\mbox{d}}x=-\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{\mbox{d}}x={\frac {1}{n-1}}\sec ^{n-2}x\tan x+{\frac {n-2}{n-1}}\int \sec ^{n-2}x{\mbox{d}}x+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{\mbox{d}}x=-{\frac {1}{n-1}}\csc ^{n-2}x\cot x+{\frac {n-2}{n-1}}\int \csc ^{n-2}x{\mbox{d}}x+C\quad \forall n\geq 2} 含有反三角函数的积分 ∫ arcsin x d x = x arcsin x + 1 − x 2 + C {\displaystyle \int \arcsin x{\mbox{d}}x=x\arcsin x+{\sqrt {1-x^{2}}}+C} ∫ arccos x d x = x arccos x − 1 − x 2 + C {\displaystyle \int \arccos x{\mbox{d}}x=x\arccos x-{\sqrt {1-x^{2}}}+C} ∫ arctan x d x = x arctan x − 1 2 ln | 1 + x 2 | + C {\displaystyle \int \arctan {x}\,dx=x\arctan {x}-{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C} ∫ arccot x d x = x arccot x + 1 2 ln | 1 + x 2 | + C {\displaystyle \int \operatorname {arccot} {x}\,dx=x\operatorname {arccot} {x}+{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C} ∫ arcsec x d x = x arcsec x − sgn ( x ) ln | x + x 2 − 1 | + C = x arcsec x + sgn ( x ) ln | x − x 2 − 1 | + C {\displaystyle \int \operatorname {arcsec} x{\mbox{d}}x=x\operatorname {arcsec} x-\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\operatorname {arcsec} x+\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C} ∫ arccsc x d x = x arccsc x + sgn ( x ) ln | x + x 2 − 1 | + C = x arccsc x − sgn ( x ) ln | x − x 2 − 1 | + C {\displaystyle \int \operatorname {arccsc} x{\mbox{d}}x=x\operatorname {arccsc} x+\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\operatorname {arccsc} x-\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C} 含有指数函数的积分 ∫ e x d x = e x + C {\displaystyle \int e^{x}{\mbox{d}}x=e^{x}+C} ∫ α x d x = α x ln α + C {\displaystyle \int \alpha ^{x}{\mbox{d}}x={\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}{\mbox{d}}x={\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}{\mbox{d}}x={\frac {1}{a}}x^{n}e^{ax}-{\frac {n}{a}}\int x^{n-1}e^{ax}{\mbox{d}}x} ∫ 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{\mbox{d}}x={\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{\mbox{d}}x={\frac {e^{ax}}{a^{2}+b^{2}}}(a\cos bx+b\sin bx)+C} 含有对数函数的积分 ∫ ln x d x = x ln x − x + C {\displaystyle \int \ln x{\mbox{d}}x=x\ln x-x+C} ∫ log α x d x = 1 ln α ( x ln x − x ) + C {\displaystyle \int \log _{\alpha }x{\mbox{d}}x={\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{\mbox{d}}x={\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}}}{\mbox{d}}x=\ln {(\ln {x})}+C} 含有双曲函数的积分 ∫ sinh x d x = cosh x + C {\displaystyle \int \sinh x{\mbox{d}}x=\cosh x+C} ∫ cosh x d x = sinh x + C {\displaystyle \int \cosh x{\mbox{d}}x=\sinh x+C} ∫ tanh x d x = ln ( cosh x ) + C {\displaystyle \int \tanh x{\mbox{d}}x=\ln \left(\cosh x\right)+C} ∫ coth x d x = ln | sinh x | + C {\displaystyle \int \coth x{\mbox{d}}x=\ln \left|\sinh x\right|+C} ∫ sech x d x = arcsin ( tanh x ) + C = arctan ( sinh x ) + C {\displaystyle \int {\mbox{sech}}\ x{\mbox{d}}x=\arcsin \left(\tanh x\right)+C=\arctan \left(\sinh x\right)+C} ∫ csch x d x = ln | tanh x 2 | + C {\displaystyle \int {\mbox{csch}}\ x{\mbox{d}}x=\ln \left|\tanh {x \over 2}\right|+C} 定积分 ∫ − ∞ ∞ e − α x 2 d x = π α {\displaystyle \int _{-\infty }^{\infty }e^{-\alpha x^{2}}{\mbox{d}}x={\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 且 n 為 奇 數 n − 1 n ⋅ n − 3 n − 2 ⋅ … ⋅ 3 4 ⋅ 1 2 ⋅ π 2 , if n > 0 且 n 為 偶 數 {\displaystyle \int _{0}^{\frac {\pi }{2}}{\mbox{sin}}^{n}x{\mbox{d}}x=\int _{0}^{\frac {\pi }{2}}{\mbox{cos}}^{n}x{\mbox{d}}x={\begin{cases}{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdot \ldots \cdot {\frac {4}{5}}\cdot {\frac {2}{3}},&{\mbox{if }}n>1{\mbox{ 且 }}n{\mbox{為 奇 數 }}\\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdot \ldots \cdot {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot {\frac {\pi }{2}},&{\mbox{if }}n>0{\mbox{ 且 }}n{\mbox{為 偶 數 }}\end{cases}}} [1]^ 这是沃利斯公式的一个情形，详见沃利斯乘积
