What is the integral of (4-3x)^5 ?

The integral of (4-3x)^5 is -1/18 (4-3x)^6+C

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integral of (4-3x)^5

\int (4 - 3 x)^{5} d x

- \frac{1}{18} (4 - 3 x)^{6} + C

Solution steps

\int (4 - 3 x)^{5} d x

Apply u-substitution

: \int - \frac{u ^{5}}{3} d u

= \int - \frac{u ^{5}}{3} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= - \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = 4 - 3 x

= - \frac{1}{3} \cdot \frac{( 4 - 3 x ) ^{6}}{6}

Simplify

- \frac{1}{3} \cdot \frac{( 4 - 3 x ) ^{6}}{6} : - \frac{1}{18} (4 - 3 x)^{6}

= - \frac{1}{18} (4 - 3 x)^{6}

Add a constant to the solution

= - \frac{1}{18} (4 - 3 x)^{6} + C

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