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

The integral of (3-4x)^9 is -1/40 (3-4x)^{10}+C

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integral of (3-4x)^9

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

- \frac{1}{40} (3 - 4 x)^{10} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u ^{9}}{4} d u

= \int - \frac{u ^{9}}{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^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= - \frac{1}{4} \cdot \frac{u ^{10}}{10}

Substitute back

u = 3 - 4 x

= - \frac{1}{4} \cdot \frac{( 3 - 4 x ) ^{10}}{10}

Simplify

- \frac{1}{4} \cdot \frac{( 3 - 4 x ) ^{10}}{10} : - \frac{1}{40} (3 - 4 x)^{10}

= - \frac{1}{40} (3 - 4 x)^{10}

Add a constant to the solution

= - \frac{1}{40} (3 - 4 x)^{10} + C

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What is the integral of (3-4x)^9 ?
The integral of (3-4x)^9 is -1/40 (3-4x)^{10}+C