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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{8}} d u

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

Apply the Power Rule

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

= - \frac{1}{7 u ^{7}}

Substitute back

u = x + 1

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

Add a constant to the solution

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

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\frac{\partial}{\partial x} (ln (x + 5 y + 6 z))

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