Answer
$$2\sec x + \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {1 + \tan x} \right)}^2}\sec x} dx \cr
& {\text{Expand the binomial}} \cr
& = \int {\left( {1 + 2\tan x + {{\tan }^2}x} \right)\sec x} dx \cr
& {\text{Multiply}} \cr
& = \int {\left( {\sec x + 2\sec x\tan x + {{\tan }^2}x\sec x} \right)} dx \cr
& {\text{Distribute}} \cr
& = \int {\sec x} dx + 2\int {\sec x\tan x} dx + \int {{{\tan }^2}x\sec x} dx \cr
& {\text{Use pythagorean identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr
& = \int {\sec x} dx + 2\int {\sec x\tan x} dx + \int {\left( {{{\sec }^2}x - 1} \right)\sec x} dx \cr
& = \int {\sec x} dx + 2\int {\sec x\tan x} dx + \int {{{\sec }^3}x} dx - \int {\sec x} dx \cr
& = 2\int {\sec x\tan x} dx + \int {{{\sec }^3}x} dx \cr
& {\text{Integrating}} \cr
& = 2\sec x + \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr} $$