Answer
$$\tan x - x + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{{\sec }^2}x - 1} \right)} dx \cr
& {\text{sum Rule}} \cr
& = \int {{{\sec }^2}x} dx - \int {dx} \cr
& {\text{integrate}} \cr
& = \tan x - x + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dx}}\left( {\tan x - x + C} \right) \cr
& = \frac{d}{{dx}}\left( {\tan x} \right) + \frac{d}{{dx}}\left( { - x} \right) + \frac{d}{{dx}}\left( C \right) \cr
& = {\sec ^2}x - 1 + 0 \cr
& = {\sec ^2}x - 1 \cr} $$
