What is the integral of (9-4x)^3 ?

The integral of (9-4x)^3 is -1/16 (9-4x)^4+C

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integral of (9-4x)^3

\int (9 - 4 x)^{3} d x

- \frac{1}{16} (9 - 4 x)^{4} + C

Solution steps

\int (9 - 4 x)^{3} d x

Apply u-substitution

: \int - \frac{u ^{3}}{4} d u

= \int - \frac{u ^{3}}{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^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = 9 - 4 x

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

Simplify

- \frac{1}{4} \cdot \frac{( 9 - 4 x ) ^{4}}{4} : - \frac{1}{16} (9 - 4 x)^{4}

= - \frac{1}{16} (9 - 4 x)^{4}

Add a constant to the solution

= - \frac{1}{16} (9 - 4 x)^{4} + C

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