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

The integral of x/(\sqrt[5]{2x^2+1)} is 5/16 (2x^2+1)^{4/5}+C

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

\frac{5}{16} (2 x^{2} + 1)^{\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}{4} \cdot \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = 2 x^{2} + 1

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

Simplify

\frac{1}{4} \cdot \frac{5}{4} (2 x^{2} + 1)^{\frac{4}{5}} : \frac{5}{16} (2 x^{2} + 1)^{\frac{4}{5}}

= \frac{5}{16} (2 x^{2} + 1)^{\frac{4}{5}}

Add a constant to the solution

= \frac{5}{16} (2 x^{2} + 1)^{\frac{4}{5}} + C

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What is the integral of x/(\sqrt[5]{2x^2+1)} ?
The integral of x/(\sqrt[5]{2x^2+1)} is 5/16 (2x^2+1)^{4/5}+C