$\int \frac{\sec x}{\sec x - \tan x} dx$ equals:

$\sec x + \tan x + c$

The correct answer is Option (2) → $\sec x + \tan x + c$

$\int \frac{\sec x}{\sec x - \tan x} dx$

$= \int \frac{\sec x (\sec x + \tan x)}{(\sec x - \tan x)(\sec x + \tan x)} dx$

$= \int \sec^2 x dx + \int \sec x \tan x dx \quad [\sec^2 x - \tan^2 x = 1]$

$= \tan x + \sec x + c$