Answer
$$\frac{1}{2}\ln \left| {\sec {x^2} + \tan {x^2}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {x\sec {x^2}} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = {x^2},\,\,\,\,du = 2xdx,\,\,\,\,dx = \frac{{du}}{{2x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {x\sec {x^2}} dx = \int {x\sec u} \left( {\frac{{du}}{{2x}}} \right) \cr
& {\text{Cancel common factor }}x \cr
& =\frac{1}{2} \int {\sec u} du \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\ln \left| {\sec u + \tan u} \right| + C \cr
& {\text{replacing }}u = {x^2} \cr
& =\frac{1}{2} \ln \left| {\sec {x^2} + \tan {x^2}} \right| + C \cr} $$
