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

The integral of (z^2)/(\sqrt[3]{4+z^3)} is 1/2 (4+z^3)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (z^2)/(\sqrt[3]{4+z^3)}

\frac{1}{2} (4 + 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{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 4 + z^{3}

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (z^2)/(\sqrt[3]{4+z^3)} ?
The integral of (z^2)/(\sqrt[3]{4+z^3)} is 1/2 (4+z^3)^{2/3}+C