Answer
$$\frac{{x\sqrt {{x^2} + 9} }}{2} - \frac{{9\ln \left( {\left| {\sqrt {{x^2} + 9} + x} \right|} \right)}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{\sqrt {9 + {x^2}} }}} dx \cr
& {\text{Substitute }}x = 3\tan \theta ,{\text{ }}dx = 3{\sec ^2}\theta d\theta \cr
& \int {\frac{{{x^2}}}{{\sqrt {9 + {x^2}} }}} dx = \int {\frac{{{{\left( {3\tan \theta } \right)}^2}}}{{\sqrt {9 + {{\left( {3\tan \theta } \right)}^2}} }}} \left( {3{{\sec }^2}\theta d\theta } \right) \cr
& {\text{Simplify}} \cr
& = \int {\frac{{9{{\tan }^2}\theta }}{{\sqrt {9 + 9{{\tan }^2}\theta } }}} \left( {3{{\sec }^2}\theta d\theta } \right) \cr
& = \int {\frac{{9{{\tan }^2}\theta }}{{\sqrt {9\left( {1 + {{\tan }^2}\theta } \right)} }}} \left( {3{{\sec }^2}\theta d\theta } \right) \cr
& = \int {\frac{{27{{\tan }^2}\theta {{\sec }^2}\theta }}{{3\sqrt {1 + {{\tan }^2}\theta } }}} d\theta \cr
& {\text{where se}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta \cr
& = 9\int {\frac{{{{\tan }^2}\theta {{\sec }^2}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} d\theta \cr
& = 9\int {{{\tan }^2}\theta \sec \theta } d\theta \cr
& = 9\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta } d\theta \cr
& {\text{multiply}} \cr
& = 9\int {\left( {{{\sec }^3}\theta - \sec \theta } \right)} d\theta \cr
& = 9\int {{{\sec }^3}\theta } d\theta - 9\int {\sec \theta } d\theta \cr
& {\text{use the formula }}\int {{{\sec }^3}\theta d\theta } = \frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\int {\sec \theta } d\theta \cr
& = 9\left( {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\int {\sec \theta } d\theta - \int {\sec \theta } d\theta } \right) \cr
& = 9\left( {\frac{1}{2}\sec \theta \tan \theta - \frac{1}{2}\int {\sec \theta } d\theta } \right) \cr
& = \frac{9}{2}\sec \theta \tan \theta - \frac{9}{2}\int {\sec \theta } d\theta \cr
& {\text{Use }}\int {\sec \theta } d\theta = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& = \frac{9}{2}\sec \theta \tan \theta - \frac{9}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& \cr
& {\text{Writing the answer in terms of }}x,{\text{ we have that }} \cr
& x = 3\tan \theta ,\,\,\,\,\sqrt {9 + {x^2}} = \sqrt {9 + {{\left( {3\tan \theta } \right)}^2}} = 3\sqrt {1 + {{\tan }^2}\theta } ,\,\,\,\,\sec \theta = \frac{{\sqrt {9 + {x^2}} }}{3} \cr
& {\text{then}}{\text{,}} \cr
& = \frac{9}{2}\left( {\frac{{\sqrt {9 + {x^2}} }}{3}} \right)\left( {\frac{x}{3}} \right) - \frac{9}{2}\ln \left| {\frac{{\sqrt {9 + {x^2}} }}{3} + \frac{x}{3}} \right| + C \cr
& = \frac{1}{2}x\sqrt {9 + {x^2}} - \frac{9}{2}\ln \left| {\frac{{\sqrt {9 + {x^2}} + x}}{3}} \right| + C \cr
& {\text{Using logarithmic properties}} \cr
& = \frac{1}{2}x\sqrt {9 + {x^2}} - \frac{9}{2}\ln \left| {\sqrt {{x^2} + 9} + x} \right| - \frac{9}{2}\ln \left| 3 \right| + C \cr
& {\text{Combining constants}} \cr
& = \frac{{x\sqrt {{x^2} + 9} }}{2} - \frac{{9\ln \left( {\left| {\sqrt {{x^2} + 9} + x} \right|} \right)}}{2} + C \cr} $$
