What is the integral of \sqrt[5]{5-2x} ?

The integral of \sqrt[5]{5-2x} is -5/12 (5-2x)^{6/5}+C

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integral of \sqrt[5]{5-2x}

\int 5 5 - 2 x d x

- \frac{5}{12} (5 - 2 x)^{\frac{6}{5}} + C

Solution steps

\int 5 5 - 2 x d x

Apply u-substitution

: \int - \frac{5 u}{2} d u

= \int - \frac{5 u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int 5 u d u

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= - \frac{1}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 5 - 2 x

= - \frac{1}{2} \cdot \frac{5}{6} (5 - 2 x)^{\frac{6}{5}}

Simplify

- \frac{1}{2} \cdot \frac{5}{6} (5 - 2 x)^{\frac{6}{5}} : - \frac{5}{12} (5 - 2 x)^{\frac{6}{5}}

= - \frac{5}{12} (5 - 2 x)^{\frac{6}{5}}

Add a constant to the solution

= - \frac{5}{12} (5 - 2 x)^{\frac{6}{5}} + C

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What is the integral of \sqrt[5]{5-2x} ?
The integral of \sqrt[5]{5-2x} is -5/12 (5-2x)^{6/5}+C