Answer
$\int\frac{9r^2 dr}{\sqrt{1-r^3}} = -6\sqrt{1-r^3}+C$
Work Step by Step
$\int\frac{9r^2 dr}{\sqrt{1-r^3}}\space$ and $\space u = 1 - r^3$
$du = -3r^2 dr$
Doing the substitution $\space u = 1-r^3$
$\int\frac{-3du}{\sqrt{u}}\space\Rightarrow\space-3\int\frac{du}{\sqrt{u}}$
Applying the integrative rules
$-3\int\frac{du}{\sqrt{u}} = -3\cdot 2\sqrt{u}+C\space\Rightarrow\space-3\int\frac{du}{\sqrt{u}} = -6\sqrt{u}+C$
Backing to $r$
$\int\frac{9r^2 dr}{\sqrt{1-r^3}} = -6\sqrt{1-r^3}+C$
