What is the integral of (x+1)^{-8}x ?

The integral of (x+1)^{-8}x is -1/(6(x+1)^6)+1/(7(x+1)^7)+C

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integral of (x+1)^{-8}x

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

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

Solution steps

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

Simplify

(x + 1)^{- 8} x : \frac{x}{( x + 1 ) ^{8}}

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

Apply u-substitution

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

Expand

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

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

Apply the Sum Rule:

\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}})

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

What is the integral of (x+1)^{-8}x ?
The integral of (x+1)^{-8}x is -1/(6(x+1)^6)+1/(7(x+1)^7)+C