What is the integral of (9-4x)^{10} ?

The integral of (9-4x)^{10} is -1/44 (9-4x)^{11}+C

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integral of (9-4x)^{10}

\int (9 - 4 x)^{10} d x

- \frac{1}{44} (9 - 4 x)^{11} + C

Solution steps

\int (9 - 4 x)^{10} d x

Apply u-substitution

: \int - \frac{u ^{10}}{4} d u

= \int - \frac{u ^{10}}{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^{10} d u

Apply the Power Rule

: \frac{u ^{11}}{11}

= - \frac{1}{4} \cdot \frac{u ^{11}}{11}

Substitute back

u = 9 - 4 x

= - \frac{1}{4} \cdot \frac{( 9 - 4 x ) ^{11}}{11}

Simplify

- \frac{1}{4} \cdot \frac{( 9 - 4 x ) ^{11}}{11} : - \frac{1}{44} (9 - 4 x)^{11}

= - \frac{1}{44} (9 - 4 x)^{11}

Add a constant to the solution

= - \frac{1}{44} (9 - 4 x)^{11} + C

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What is the integral of (9-4x)^{10} ?
The integral of (9-4x)^{10} is -1/44 (9-4x)^{11}+C