What is the integral of \sqrt[5]{(8-3x)^6} ?

The integral of \sqrt[5]{(8-3x)^6} is -5/33 (8-3x)^{11/5}+C

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integral of \sqrt[5]{(8-3x)^6}

- \frac{5}{33} (8 - 3 x)^{\frac{11}{5}} + C

Solution steps

Apply u-substitution

: \int - \frac{u ^{\frac{6}{5}}}{3} d u

= \int - \frac{u ^{\frac{6}{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^{\frac{6}{5}} d u

Apply the Power Rule

: \frac{5}{11} u^{\frac{11}{5}}

= - \frac{1}{3} \cdot \frac{5}{11} u^{\frac{11}{5}}

Substitute back

u = 8 - 3 x

= - \frac{1}{3} \cdot \frac{5}{11} (8 - 3 x)^{\frac{11}{5}}

Simplify

- \frac{1}{3} \cdot \frac{5}{11} (8 - 3 x)^{\frac{11}{5}} : - \frac{5}{33} (8 - 3 x)^{\frac{11}{5}}

= - \frac{5}{33} (8 - 3 x)^{\frac{11}{5}}

Add a constant to the solution

= - \frac{5}{33} (8 - 3 x)^{\frac{11}{5}} + C

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What is the integral of \sqrt[5]{(8-3x)^6} ?
The integral of \sqrt[5]{(8-3x)^6} is -5/33 (8-3x)^{11/5}+C