Answer
$$\frac{9}{2}{\sin ^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{2}x\sqrt {9 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{\sqrt {9 - {x^2}} }}} dx \cr
& {\text{Substitute }}x = 3\sin \theta ,{\text{ }}dx = 3\cos \theta d\theta \cr
& {\text{Use the trigonometric substitution}} \cr
& \int {\frac{{{x^2}}}{{\sqrt {9 - {x^2}} }}} dx = \int {\frac{{{{\left( {3\sin \theta } \right)}^2}}}{{\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \left( {3\cos \theta d\theta } \right) \cr
& {\text{Simplify}} \cr
& = \int {\frac{{9{{\sin }^2}\theta }}{{\sqrt {9 - 9{{\sin }^2}\theta } }}} \left( {3\cos \theta d\theta } \right) \cr
& = \int {\frac{{27{{\sin }^2}\theta \cos \theta d\theta }}{{3\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& = \int {\frac{{9{{\sin }^2}\theta \cos \theta d\theta }}{{\cos \theta }}} \cr
& = 9\int {{{\sin }^2}\theta } d\theta \cr
& {\text{Use the identity }}{\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2} \cr
& = 9\int {\left( {\frac{{1 - \cos 2\theta }}{2}} \right)} d\theta \cr
& = \frac{9}{2}\int {\left( {1 - \cos 2\theta } \right)} d\theta \cr
& {\text{Integrate}} \cr
& = \frac{9}{2}\left( {\theta - \frac{1}{2}\sin 2\theta } \right) + C \cr
& = \frac{9}{2}\theta - \frac{9}{4}\sin 2\theta + C \cr
& = \frac{9}{2}\theta - \frac{9}{4}\left( {2\sin \theta \cos \theta } \right) + C \cr
& = \frac{9}{2}\theta - \frac{9}{2}\sin \theta \cos \theta + C \cr
& {\text{Writing the answer in terms of }}x,{\text{ we have that: }} \cr
& x = 3\sin \theta ,{\text{so }}\sin \theta = \frac{x}{3},{\text{ and cos}}\theta = \frac{{\sqrt {9 - {x^2}} }}{3} \cr
& = \frac{9}{2}{\sin ^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{9}{2}\left( {\frac{x}{3}} \right)\left( {\frac{{\sqrt {9 - {x^2}} }}{3}} \right) + C \cr
& = \frac{9}{2}{\sin ^{ - 1}}\left( {\frac{x}{3}} \right) - \frac{1}{2}x\sqrt {9 - {x^2}} + C \cr} $$
