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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{z}{5 z ^{2} + 2} d z

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{5 u} d u

Apply the Power Rule

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

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

Substitute back

u = z^{2} + 2

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

Simplify

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

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

Add a constant to the solution

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

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