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

The integral of x/(\sqrt[5]{2x^2+4)} is 5/(8\sqrt[5]{2)}(x^2+2)^{4/5}+C

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

\int \frac{x}{5 2 x ^{2} + 4} d x

\frac{5}{8 5 2} (x^{2} + 2)^{\frac{4}{5}} + C

Solution steps

\int \frac{x}{5 2 x ^{2} + 4} d x

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

= \int \frac{1}{5 2} \cdot \frac{x}{5 x ^{2} + 2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5 2} \cdot \int \frac{x}{5 x ^{2} + 2} d x

Apply u-substitution

: \int \frac{1}{2 5 u} d u

= \frac{1}{5 2} \cdot \int \frac{1}{2 5 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5 2} \cdot \frac{1}{2} \cdot \int \frac{1}{5 u} d u

Apply the Power Rule

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

= \frac{1}{5 2} \cdot \frac{1}{2} \cdot \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = x^{2} + 2

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

Simplify

\frac{1}{5 2} \cdot \frac{1}{2} \cdot \frac{5}{4} (x^{2} + 2)^{\frac{4}{5}} : \frac{5}{8 5 2} (x^{2} + 2)^{\frac{4}{5}}

= \frac{5}{8 5 2} (x^{2} + 2)^{\frac{4}{5}}

Add a constant to the solution

= \frac{5}{8 5 2} (x^{2} + 2)^{\frac{4}{5}} + C

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What is the integral of x/(\sqrt[5]{2x^2+4)} ?
The integral of x/(\sqrt[5]{2x^2+4)} is 5/(8\sqrt[5]{2)}(x^2+2)^{4/5}+C