What is the integral of 3/(\sqrt[5]{x)} ?

The integral of 3/(\sqrt[5]{x)} is 15/4 x^{4/5}+C

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integral of 3/(\sqrt[5]{x)}

\frac{15}{4} x^{\frac{4}{5}} + C

Solution steps

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} x^{\frac{4}{5}}

= 3 \cdot \frac{5}{4} x^{\frac{4}{5}}

Simplify

3 \cdot \frac{5}{4} x^{\frac{4}{5}} : \frac{15}{4} x^{\frac{4}{5}}

= \frac{15}{4} x^{\frac{4}{5}}

Add a constant to the solution

= \frac{15}{4} x^{\frac{4}{5}} + C

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What is the integral of 3/(\sqrt[5]{x)} ?
The integral of 3/(\sqrt[5]{x)} is 15/4 x^{4/5}+C