Answer
$\frac{(1+z^3)^{2/3}}{2} +C$
Work Step by Step
Evaluate the Integral using substitution: $\int \frac{z^2}{\sqrt[3] {1+z^3}}dz$
Substitution Rule: $\int f(g(x))g’(x)dx = \int f(u)du$
$u= 1+z^3$
$du =3z^2$
Since $du$ in the expression is equal to $(z^2)$ it must be multiplied by $\frac{1}{3}$
Solve the integral in terms of $u$:
$\int \frac{1}{\sqrt[3] u}(\frac13)du$
$\frac{1}{3}\int u^{-1/3}du $
$\frac{1}{3}\frac{3u^{2/3}}{2} +C$
$\frac{u^{2/3}}{2} + C$
Substitute for $u$:
$\frac{(1+z^3)^{2/3}}{2} +C$
