What is the integral of (4-3x)^6 ?

The integral of (4-3x)^6 is -1/21 (4-3x)^7+C

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integral of (4-3x)^6

\int (4 - 3 x)^{6} d x

- \frac{1}{21} (4 - 3 x)^{7} + C

Solution steps

\int (4 - 3 x)^{6} d x

Apply u-substitution

= \int - \frac{u ^{6}}{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^{6} d u

Apply the Power Rule

= - \frac{1}{3} \cdot \frac{u ^{7}}{7}

Substitute back

u = 4 - 3 x

= - \frac{1}{3} \cdot \frac{( 4 - 3 x ) ^{7}}{7}

Simplify

- \frac{1}{3} \cdot \frac{( 4 - 3 x ) ^{7}}{7} : - \frac{1}{21} (4 - 3 x)^{7}

= - \frac{1}{21} (4 - 3 x)^{7}

Add a constant to the solution

= - \frac{1}{21} (4 - 3 x)^{7} + C