What is the integral of \sqrt[5]{1-3x} ?

The integral of \sqrt[5]{1-3x} is -5/18 (1-3x)^{6/5}+C

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integral of \sqrt[5]{1-3x}

- \frac{5}{18} (1 - 3 x)^{\frac{6}{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}{6} u^{\frac{6}{5}}

= - \frac{1}{3} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 1 - 3 x

= - \frac{1}{3} \cdot \frac{5}{6} (1 - 3 x)^{\frac{6}{5}}

Simplify

- \frac{1}{3} \cdot \frac{5}{6} (1 - 3 x)^{\frac{6}{5}} : - \frac{5}{18} (1 - 3 x)^{\frac{6}{5}}

= - \frac{5}{18} (1 - 3 x)^{\frac{6}{5}}

Add a constant to the solution

= - \frac{5}{18} (1 - 3 x)^{\frac{6}{5}} + C

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What is the integral of \sqrt[5]{1-3x} ?
The integral of \sqrt[5]{1-3x} is -5/18 (1-3x)^{6/5}+C