What is the integral of (z^3)/(\sqrt[3]{6+z^4)} ?

The integral of (z^3)/(\sqrt[3]{6+z^4)} is 3/8 (6+z^4)^{2/3}+C

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integral of (z^3)/(\sqrt[3]{6+z^4)}

\frac{3}{8} (6 + z^{4})^{\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}{4} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 6 + z^{4}

= \frac{1}{4} \cdot \frac{3}{2} (6 + z^{4})^{\frac{2}{3}}

Simplify

\frac{1}{4} \cdot \frac{3}{2} (6 + z^{4})^{\frac{2}{3}} : \frac{3}{8} (6 + z^{4})^{\frac{2}{3}}

= \frac{3}{8} (6 + z^{4})^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{8} (6 + z^{4})^{\frac{2}{3}} + C

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What is the integral of (z^3)/(\sqrt[3]{6+z^4)} ?
The integral of (z^3)/(\sqrt[3]{6+z^4)} is 3/8 (6+z^4)^{2/3}+C