What is the integral of (z^2)/(\sqrt[3]{9+z^3)} ?

The integral of (z^2)/(\sqrt[3]{9+z^3)} is 1/2 (9+z^3)^{2/3}+C

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integral of (z^2)/(\sqrt[3]{9+z^3)}

\frac{1}{2} (9 + z^{3})^{\frac{2}{3}} + 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{3}{2} u^{\frac{2}{3}}

= \frac{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 9 + z^{3}

= \frac{1}{3} \cdot \frac{3}{2} (9 + z^{3})^{\frac{2}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{2} (9 + z^{3})^{\frac{2}{3}} : \frac{1}{2} (9 + z^{3})^{\frac{2}{3}}

= \frac{1}{2} (9 + z^{3})^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{2} (9 + z^{3})^{\frac{2}{3}} + C

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What is the integral of (z^2)/(\sqrt[3]{9+z^3)} ?
The integral of (z^2)/(\sqrt[3]{9+z^3)} is 1/2 (9+z^3)^{2/3}+C