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

The integral of (x-1)^9 is x(x-1)^9-9/10 (x-1)^{10}-(x-1)^9+C

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integral of (x-1)^9

\int (x - 1)^{9} d x

x (x - 1)^{9} - \frac{9}{10} (x - 1)^{10} - (x - 1)^{9} + C

Solution steps

\int (x - 1)^{9} d x

Apply Integration By Parts

: (x - 1)^{9} x - \int 9 (x - 1)^{8} x d x

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

\int 9 (x - 1)^{8} x d x = \frac{9}{10} (x - 1)^{10} + (x - 1)^{9}

= (x - 1)^{9} x - (\frac{9}{10} (x - 1)^{10} + (x - 1)^{9})

Simplify

(x - 1)^{9} x - (\frac{9}{10} (x - 1)^{10} + (x - 1)^{9}) : x (x - 1)^{9} - \frac{9}{10} (x - 1)^{10} - (x - 1)^{9}

= x (x - 1)^{9} - \frac{9}{10} (x - 1)^{10} - (x - 1)^{9}

Add a constant to the solution

= x (x - 1)^{9} - \frac{9}{10} (x - 1)^{10} - (x - 1)^{9} + C

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What is the integral of (x-1)^9 ?
The integral of (x-1)^9 is x(x-1)^9-9/10 (x-1)^{10}-(x-1)^9+C