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

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

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integral of (x+1)^8

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

\frac{1}{9} x (x + 1)^{8} + \frac{1}{9} (x + 1)^{8} + C

Solution steps

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

Apply Integration By Parts

: (x + 1)^{8} x - \int 8 (x + 1)^{7} x d x

= (x + 1)^{8} x - \int 8 (x + 1)^{7} x d x

\int 8 (x + 1)^{7} x d x = \frac{1}{9} (x + 1)^{8} (8 x - 1)

= (x + 1)^{8} x - \frac{1}{9} (x + 1)^{8} (8 x - 1)

Simplify

(x + 1)^{8} x - \frac{1}{9} (x + 1)^{8} (8 x - 1) : \frac{1}{9} x (x + 1)^{8} + \frac{1}{9} (x + 1)^{8}

= \frac{1}{9} x (x + 1)^{8} + \frac{1}{9} (x + 1)^{8}

Add a constant to the solution

= \frac{1}{9} x (x + 1)^{8} + \frac{1}{9} (x + 1)^{8} + C

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