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

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

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

\int \frac{1}{5 x + 2} d x

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

Solution steps

\int \frac{1}{5 x + 2} d x

Apply u-substitution

= \int \frac{1}{5 u} d u

Apply the Power Rule

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

Substitute back

u = x + 2

= \frac{5}{4} (x + 2)^{\frac{4}{5}}

Add a constant to the solution

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

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