What is the integral of (9-4x)^8 ?

The integral of (9-4x)^8 is -1/36 (9-4x)^9+C

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integral of (9-4x)^8

\int (9 - 4 x)^{8} d x

- \frac{1}{36} (9 - 4 x)^{9} + C

Solution steps

\int (9 - 4 x)^{8} d x

Apply u-substitution

= \int - \frac{u ^{8}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int u^{8} d u

Apply the Power Rule

= - \frac{1}{4} \cdot \frac{u ^{9}}{9}

Substitute back

u = 9 - 4 x

= - \frac{1}{4} \cdot \frac{( 9 - 4 x ) ^{9}}{9}

Simplify

- \frac{1}{4} \cdot \frac{( 9 - 4 x ) ^{9}}{9} : - \frac{1}{36} (9 - 4 x)^{9}

= - \frac{1}{36} (9 - 4 x)^{9}

Add a constant to the solution

= - \frac{1}{36} (9 - 4 x)^{9} + C