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