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