Answer
$x\sec x - \ln \left| {\sec x + \tan x} \right| + C$
Work Step by Step
$$\eqalign{
& \int {x\sec x\tan x} dx \cr
& {\text{Integrate by parts }} \cr
& u = x,{\text{ }}du = dx \cr
& dv = \sec x\tan xdx,{\text{ }}v = \sec x \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {x\sec x\tan x} dx = x\sec x - \int {\sec x} dx \cr
& \int {x\sec x\tan x} dx = x\sec x - \ln \left| {\sec x + \tan x} \right| + C \cr} $$
