Answer
$$\frac{1}{2}x\sqrt {{x^2} + 1} - \frac{1}{2}\ln \left| {\sqrt {{x^2} + 1} + x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} dx \cr
& {\text{Let }}x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} dx = \int {\frac{{{{\left( {\tan \theta } \right)}^2}}}{{\sqrt {{{\left( {\tan \theta } \right)}^2} + 1} }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{{{\tan }^2}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = \int {{{\tan }^2}\theta \sec \theta } d\theta \cr
& {\text{Use pythagorean identity}} \cr
& = \int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta } d\theta \cr
& = \int {\left( {{{\sec }^3}\theta - \sec \theta } \right)} d\theta \cr
& = \int {{{\sec }^3}\theta } d\theta - \int {\sec \theta } d\theta \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| - \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& = \frac{1}{2}\sec \theta \tan \theta - \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{2}x\sqrt {{x^2} + 1} - \frac{1}{2}\ln \left| {\sqrt {{x^2} + 1} + x} \right| + C \cr} $$
