integral of (e^x(e^x+1))/((e^x-1)^3)

解答

\int \frac{e ^{x} ( e ^{x} + 1 )}{( e ^{x} - 1 ) ^{3}} d x

- \frac{1}{e ^{x} - 1} - \frac{1}{( e ^{x} - 1 ) ^{2}} + C

求解步骤

\int \frac{e ^{x} ( e ^{x} + 1 )}{( e ^{x} - 1 ) ^{3}} d x

使用换元积分法

= \int \frac{u + 1}{( u - 1 ) ^{3}} d u

乘开

\frac{u + 1}{( u - 1 ) ^{3}} : \frac{u}{( u - 1 ) ^{3}} + \frac{1}{( u - 1 ) ^{3}}

= \int \frac{u}{( u - 1 ) ^{3}} + \frac{1}{( u - 1 ) ^{3}} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{( u - 1 ) ^{3}} d u + \int \frac{1}{( u - 1 ) ^{3}} d u

\int \frac{u}{( u - 1 ) ^{3}} d u = - \frac{1}{u - 1} - \frac{1}{2 ( u - 1 ) ^{2}}

\int \frac{1}{( u - 1 ) ^{3}} d u = - \frac{1}{2 ( u - 1 ) ^{2}}

= - \frac{1}{u - 1} - \frac{1}{2 ( u - 1 ) ^{2}} - \frac{1}{2 ( u - 1 ) ^{2}}

u = e^{x}

代回

= - \frac{1}{e ^{x} - 1} - \frac{1}{2 ( e ^{x} - 1 ) ^{2}} - \frac{1}{2 ( e ^{x} - 1 ) ^{2}}

化简

- \frac{1}{e ^{x} - 1} - \frac{1}{2 ( e ^{x} - 1 ) ^{2}} - \frac{1}{2 ( e ^{x} - 1 ) ^{2}} : - \frac{1}{e ^{x} - 1} - \frac{1}{( e ^{x} - 1 ) ^{2}}

= - \frac{1}{e ^{x} - 1} - \frac{1}{( e ^{x} - 1 ) ^{2}}

解答补常数

= - \frac{1}{e ^{x} - 1} - \frac{1}{( e ^{x} - 1 ) ^{2}} + C