Answer
$$2\ln \left| {\sec \sqrt x + \tan \sqrt x } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sec \left( {\sqrt x } \right)}}{{\sqrt x }}} dx \cr
& = \int {\frac{{\sec {x^{1/2}}}}{{{x^{1/2}}}}} dx \cr
& = \int {\sec {x^{1/2}}\left( {{x^{ - 1/2}}} \right)} dx \cr
& {\text{substitute }}u = {x^{1/2}},{\text{ }}du = \frac{1}{2}{x^{ - 1/2}}dx \cr
& 2du = {x^{ - 1/2}}dx \cr
& = \int {\sec {x^{1/2}}\left( {{x^{ - 1/2}}} \right)} dx \cr
& = \int {\sec u\left( {2du} \right)} \cr
& = 2\int {\sec udu} \cr
& {\text{find the antiderivative }}\left( {{\text{use formula 22 page 503}}} \right) \cr
& = 2\ln \left| {\sec u + \tan u} \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {x^{1/2}} \cr
& = 2\ln \left| {\sec {x^{1/2}} + \tan {x^{1/2}}} \right| + C \cr
& = 2\ln \left| {\sec \sqrt x + \tan \sqrt x } \right| + C \cr} $$
