What is the integral of 1/(\sqrt[5]{1-2x)} ?

The integral of 1/(\sqrt[5]{1-2x)} is -5/8 (1-2x)^{4/5}+C

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integral of 1/(\sqrt[5]{1-2x)}

- \frac{5}{8} (1 - 2 x)^{\frac{4}{5}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{5}{4} u^{\frac{4}{5}}

= - \frac{1}{2} \cdot \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = 1 - 2 x

= - \frac{1}{2} \cdot \frac{5}{4} (1 - 2 x)^{\frac{4}{5}}

Simplify

- \frac{1}{2} \cdot \frac{5}{4} (1 - 2 x)^{\frac{4}{5}} : - \frac{5}{8} (1 - 2 x)^{\frac{4}{5}}

= - \frac{5}{8} (1 - 2 x)^{\frac{4}{5}}

Add a constant to the solution

= - \frac{5}{8} (1 - 2 x)^{\frac{4}{5}} + C

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What is the integral of 1/(\sqrt[5]{1-2x)} ?
The integral of 1/(\sqrt[5]{1-2x)} is -5/8 (1-2x)^{4/5}+C