Answer
$$2\sqrt {\ln \left( {\sec x + \tan x} \right)} $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sec xdx}}{{\sqrt {\ln \left( {\sec x + \tan x} \right)} }}} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \ln \left( {\sec x + \tan x} \right),\cr
& {\text{ so that }}du = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}}dx \cr
& du = \sec xdx \cr
& {\text{Write the integral in terms of }}u \cr
& \int {\frac{{\sec xdx}}{{\sqrt {\ln \left( {\sec x + \tan x} \right)} }}} = \int {\frac{{du}}{{\sqrt u }}} \cr
& = \int {{u^{ - 1/2}}} du \cr
& {\text{Integrate by using the power rule}} \cr
& = \frac{{{u^{1/2}}}}{{1/2}} + C \cr
& = 2{u^{1/2}} + C \cr
& = 2\sqrt u \cr
& {\text{Write in terms of }}x:\cr
& {\text{Replace }}\ln \left( {\sec x + \tan x} \right)\tan t{\text{ for }}u \cr
& = 2\sqrt {\ln \left( {\sec x + \tan x} \right)} \cr} $$
