Answer
$$\frac{{{{\tan }^3}\left( {{x^2}} \right)}}{6} + C$$
Work Step by Step
$$\eqalign{
& \int {x{{\tan }^2}\left( {{x^2}} \right)} {\sec ^2}\left( {{x^2}} \right)dx \cr
& {\text{substitute }}u = \tan \left( {{x^2}} \right),{\text{ }}du = {\sec ^2}\left( {{x^2}} \right)\left( {2x} \right)dx \cr
& {\text{ }}\frac{1}{2}du = {\sec ^2}\left( {{x^2}} \right)xdx \cr
& \int {x{{\tan }^2}\left( {{x^2}} \right)} {\sec ^2}\left( {{x^2}} \right)dx \cr
& = \int {{u^2}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {{u^2}} du \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{2}\left( {\frac{{{u^3}}}{3}} \right) + C \cr
& = \frac{{{u^3}}}{6} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{{{{\tan }^3}\left( {{x^2}} \right)}}{6} + C \cr} $$
