Answer
$$2\tan x + 2\sec x - x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{1 + \sin x}}{{1 - \sin x}}} dx \cr
& {\text{Multiply the denominator by the conjugate of the numerator}} \cr
& = \int {\frac{{1 + \sin x}}{{1 - \sin x}}\left( {\frac{{1 + \sin x}}{{1 + \sin x}}} \right)} dx \cr
& = \int {\frac{{{{\left( {1 + \sin x} \right)}^2}}}{{1 - {{\sin }^2}x}}} dx \cr
& {\text{Expand and simplify}} \cr
& = \int {\frac{{1 + 2\sin x + {{\sin }^2}x}}{{{{\cos }^2}x}}} dx \cr
& = \int {\left( {\frac{1}{{{{\cos }^2}x}} + \frac{{2\sin x}}{{{{\cos }^2}x}} + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)} dx \cr
& = \int {\left( {{{\sec }^2}x + \frac{{2\sin x}}{{{{\left( {\cos x} \right)}^2}}} + {{\tan }^2}x} \right)} dx \cr
& = \int {\left( {{{\sec }^2}x + \frac{{2\sin x}}{{{{\left( {\cos x} \right)}^2}}} + {{\sec }^2}x - 1} \right)} dx \cr
& = \int {\left( {2{{\sec }^2}x + 2\frac{{\sin x}}{{\cos x}}\left( {\frac{1}{{\cos x}}} \right) - 1} \right)} dx \cr
& = \int {\left( {2{{\sec }^2}x + 2\sec x\tan x - 1} \right)} dx \cr
& {\text{Integrating}} \cr
& = 2\tan x + 2\sec x - x + C \cr} $$
