What is the integral of (x^2+x)^8(2x+1) ?

The integral of (x^2+x)^8(2x+1) is ((x^2+x)^9)/9+C

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integral of (x^2+x)^8(2x+1)

\int (x^{2} + x)^{8} (2 x + 1) d x

\frac{( x ^{2} + x ) ^{9}}{9} + C

Solution steps

\int (x^{2} + x)^{8} (2 x + 1) d x

Apply u-substitution

: \int u^{8} d u

= \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{u ^{9}}{9}

Substitute back

u = x^{2} + x

= \frac{( x ^{2} + x ) ^{9}}{9}

Add a constant to the solution

= \frac{( x ^{2} + x ) ^{9}}{9} + C

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What is the integral of (x^2+x)^8(2x+1) ?
The integral of (x^2+x)^8(2x+1) is ((x^2+x)^9)/9+C