What is the integral of \sqrt[5]{3y-8} ?

The integral of \sqrt[5]{3y-8} is 5/18 (3y-8)^{6/5}+C

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integral of \sqrt[5]{3y-8}

\frac{5}{18} (3 y - 8)^{\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{1}{3} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 3 y - 8

= \frac{1}{3} \cdot \frac{5}{6} (3 y - 8)^{\frac{6}{5}}

Simplify

\frac{1}{3} \cdot \frac{5}{6} (3 y - 8)^{\frac{6}{5}} : \frac{5}{18} (3 y - 8)^{\frac{6}{5}}

= \frac{5}{18} (3 y - 8)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{18} (3 y - 8)^{\frac{6}{5}} + C

What is the integral of \sqrt[5]{3y-8} ?
The integral of \sqrt[5]{3y-8} is 5/18 (3y-8)^{6/5}+C