Answer
$$\frac{1}{2}{x^2} + \frac{1}{2}\cot {x^2} + \frac{1}{2}\csc {x^2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{1 - \sec {x^2}}}} dx \cr
& {\text{Let }}u = {x^2}.{\text{ Then }}du = 2xdx,{\text{ }}dx = \frac{{du}}{{2x}}{\text{ substituting you have}} \cr
& \int {\frac{x}{{1 - \sec {x^2}}}} dx = \int {\frac{x}{{1 - \sec u}}\left( {\frac{1}{{2x}}} \right)} du \cr
& = \int {\frac{1}{{1 - \sec {u^2}}}\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\frac{1}{{1 - \sec u}}} du \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{{du}}{{1 - \sec u}}} = u + \cot u + \csc u + C \cr
& \frac{1}{2}\int {\frac{1}{{1 - \sec u}}} du = \frac{1}{2}u + \frac{1}{2}\cot u + \frac{1}{2}\csc u + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = {x^2} \cr
& = \frac{1}{2}{x^2} + \frac{1}{2}\cot {x^2} + \frac{1}{2}\csc {x^2} + C \cr} $$
