What is the integral of 1/((x+1)^9) ?

The integral of 1/((x+1)^9) is -1/(8(x+1)^8)+C

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integral of 1/((x+1)^9)

\int \frac{1}{( x + 1 ) ^{9}} d x

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

Solution steps

\int \frac{1}{( x + 1 ) ^{9}} d x

Apply u-substitution

: \int \frac{1}{u ^{9}} d u

= \int \frac{1}{u ^{9}} d u

Apply the Power Rule

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

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

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What is the integral of 1/((x+1)^9) ?
The integral of 1/((x+1)^9) is -1/(8(x+1)^8)+C