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

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

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

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

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

Solution steps

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

Simplify

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

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

Expand

\frac{x + 1}{x ^{8}} : \frac{1}{x ^{7}} + \frac{1}{x ^{8}}

= \int \frac{1}{x ^{7}} + \frac{1}{x ^{8}} d x

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}{x ^{7}} d x + \int \frac{1}{x ^{8}} d x

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

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

= - \frac{1}{6 x ^{6}} - \frac{1}{7 x ^{7}}

Add a constant to the solution

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

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