integral of x/((x+1)^8)

解答

\int \frac{x}{( x + 1 ) ^{8}} d x

- \frac{1}{6 ( x + 1 ) ^{6}} + \frac{1}{7 ( x + 1 ) ^{7}} + C

求解步骤

\int \frac{x}{( x + 1 ) ^{8}} d x

使用换元积分法

= \int \frac{u - 1}{u ^{8}} d u

乘开

\frac{u - 1}{u ^{8}} : \frac{1}{u ^{7}} - \frac{1}{u ^{8}}

= \int \frac{1}{u ^{7}} - \frac{1}{u ^{8}} 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{1}{u ^{7}} d u - \int \frac{1}{u ^{8}} d u

\int \frac{1}{u ^{7}} d u = - \frac{1}{6 u ^{6}}

\int \frac{1}{u ^{8}} d u = - \frac{1}{7 u ^{7}}

= - \frac{1}{6 u ^{6}} - (- \frac{1}{7 u ^{7}})

u = x + 1

代回

= - \frac{1}{6 ( x + 1 ) ^{6}} - (- \frac{1}{7 ( x + 1 ) ^{7}})

化简

= - \frac{1}{6 ( x + 1 ) ^{6}} + \frac{1}{7 ( x + 1 ) ^{7}}

解答补常数

= - \frac{1}{6 ( x + 1 ) ^{6}} + \frac{1}{7 ( x + 1 ) ^{7}} + C

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