Answer
$$\int\frac{9r^2dr}{\sqrt{1-r^3}}=-6\sqrt{1-r^3}+C$$
Work Step by Step
$$A=\int\frac{9r^2dr}{\sqrt{1-r^3}}$$
We set $u=1-r^3$.
Then $$du=-3r^2dr=\frac{-3}{9}\times9r^2dr=-\frac{1}{3}\times9r^2dr$$
That means $$9r^2dr=-3du$$
Therefore, $$A=-3\int\frac{du}{\sqrt u}=-3\int u^{-1/2}du$$ $$A=\frac{-3u^{1/2}}{\frac{1}{2}}+C=-6\sqrt u+C$$ $$A=-6\sqrt{1-r^3}+C$$
