Answer
$$x\operatorname{arcsec} 2x - \frac{1}{2}\ln \left| {2x + \sqrt {4{x^2} - 1} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\operatorname{arcsec} 2x} dx,{\text{ }}x > \frac{1}{2} \cr
& {\text{Integrating by parts,}} \cr
& {\text{Let }}u = \operatorname{arcsec} 2x,{\text{ }}dv = dx,{\text{ }}v = x \cr
& du = \frac{d}{{dx}}\left[ {\operatorname{arcsec} 2x} \right] = \frac{2}{{2x\sqrt {{{\left( {2x} \right)}^2} - 1} }}dx = \frac{1}{{x\sqrt {{{\left( {2x} \right)}^2} - 1} }}dx \cr
& {\text{Using the integration by parts formula}} \cr
& \int {\operatorname{arcsec} 2x} dx = x\operatorname{arcsec} 2x - \int {\frac{x}{{x\sqrt {{{\left( {2x} \right)}^2} - 1} }}} dx \cr
& = x\operatorname{arcsec} 2x - \int {\frac{1}{{\sqrt {{{\left( {2x} \right)}^2} - 1} }}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{*Integrating }}\int {\frac{1}{{\sqrt {{{\left( {2x} \right)}^2} - 1} }}} dx \cr
& {\text{Using the substitution sec}}\theta = 2x,{\text{ }}\sec \theta \tan \theta d\theta = 2dx \cr
& \int {\frac{1}{{\sqrt {{{\left( {2x} \right)}^2} - 1} }}} dx = \int {\frac{1}{{\sqrt {{{\sec }^2}\theta - 1} }}\left( {\frac{1}{2}\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\frac{1}{{\sqrt {{{\sec }^2}\theta - 1} }}\left( {\frac{1}{2}\sec \theta \tan \theta } \right)} d\theta \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt {{{\tan }^2}\theta } }}\left( {\sec \theta \tan \theta } \right)} d\theta \cr
& = \frac{1}{2}\int {\sec \theta } d\theta \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{We know that sec}}\theta = \frac{{2x}}{1},{\text{ then tan}}\theta = \frac{{\sqrt {{{\left( {2x} \right)}^2} - 1} }}{1} \cr
& {\text{ = }}\frac{1}{2}\ln \left| {2x + \sqrt {4{x^2} - 1} } \right| + C \cr
& {\text{Substituting into }}\left( {\bf{1}} \right) \cr
& = x\operatorname{arcsec} 2x - \frac{1}{2}\ln \left| {2x + \sqrt {4{x^2} - 1} } \right| + C \cr} $$