Answer
$$\frac{1}{3}{\left( {1 + {x^2}} \right)^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& \int x \sqrt {1 + {x^2}} dx \cr
& {\text{Using the substitution }}x = \tan \theta \cr
& x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{Substitute}} \cr
& \int x \sqrt {1 + {x^2}} dx = \int {\tan \theta } \sqrt {1 + {{\tan }^2}\theta } \left( {{{\sec }^2}\theta d\theta } \right) \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\tan \theta } \sqrt {{{\sec }^2}\theta } \left( {{{\sec }^2}\theta d\theta } \right) \cr
& = \int {\tan \theta } {\sec ^3}\theta d\theta \cr
& = \int {{{\sec }^2}\theta } \sec \theta \tan \theta d\theta \cr
& {\text{Integrating}} \cr
& = \frac{{{{\sec }^3}\theta }}{3} + C \cr
& {\text{Where }}\tan \theta = x,{\text{ }}\sec \theta = \sqrt {1 + {x^2}} \cr
& = \frac{1}{3}{\left( {\sqrt {1 + {x^2}} } \right)^3} + C \cr
& = \frac{1}{3}{\left( {1 + {x^2}} \right)^{3/2}} + C \cr} $$
