Answer
$$\left( {\tan x + 2} \right)\ln \left( {\tan x + 2} \right) - \tan x + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{{\sec }^2}x} \right)\ln \left( {\tan x + 2} \right)} dx \cr
& \int {\ln \left( {\tan x + 2} \right)\left( {{{\sec }^2}x} \right)} dx \cr
& {\text{Let }}t = \tan x + 2,{\text{ }}dt = {\sec ^2}xdx \cr
& \cr
& {\text{Apply the substitution}} \cr
& \int {\ln \left( {\tan x + 2} \right)\left( {{{\sec }^2}x} \right)} dx = \int {\ln t} dt \cr
& \cr
& {\text{Use the Integral of }}\ln x\left( {{\text{See page 519}}} \right):{\text{ }} \cr
& \int {\ln xdx} = x\ln x - x + C,{\text{ then}} \cr
& \int {\ln t} dt = t\ln t - t + C \cr
& {\text{Back - substitute }}t = \tan x + 2 \cr
& = \left( {\tan x + 2} \right)\ln \left( {\tan x + 2} \right) - \left( {\tan x + 2} \right) + C \cr
& = \left( {\tan x + 2} \right)\ln \left( {\tan x + 2} \right) - \tan x - 2 + C \cr
& {\text{Combine constants}} \cr
& = \left( {\tan x + 2} \right)\ln \left( {\tan x + 2} \right) - \tan x + C \cr} $$
