What is the integral of (z^4)/(\sqrt[3]{7+z^5)} ?

The integral of (z^4)/(\sqrt[3]{7+z^5)} is 3/10 (7+z^5)^{2/3}+C

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integral of (z^4)/(\sqrt[3]{7+z^5)}

\frac{3}{10} (7 + z^{5})^{\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{1}{5} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 7 + z^{5}

= \frac{1}{5} \cdot \frac{3}{2} (7 + z^{5})^{\frac{2}{3}}

Simplify

\frac{1}{5} \cdot \frac{3}{2} (7 + z^{5})^{\frac{2}{3}} : \frac{3}{10} (7 + z^{5})^{\frac{2}{3}}

= \frac{3}{10} (7 + z^{5})^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{10} (7 + z^{5})^{\frac{2}{3}} + C

What is the integral of (z^4)/(\sqrt[3]{7+z^5)} ?
The integral of (z^4)/(\sqrt[3]{7+z^5)} is 3/10 (7+z^5)^{2/3}+C