Answer
$$2{\tan ^2}x - 2\ln \left( {1 + {{\tan }^2}x} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {4{{\tan }^3}x} dx \cr
& {\text{We write }}{\tan ^3}x{\text{ as ta}}{{\text{n}}^2}x\tan x \cr
& = 4\int {{\text{ta}}{{\text{n}}^2}x\tan x} dx \cr
& {\text{use the fundamental identiy ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr
& = 4\int {\left( {{{\sec }^2}x - 1} \right)\tan x} dx \cr
& = 4\int {\left( {\tan x{{\sec }^2}x - \tan x} \right)} dx \cr
& = 4\int {\tan x{{\sec }^2}x} dx - 4\int {\tan x} dx \cr
& {\text{integrating}} \cr
& = 4\left( {\frac{{{{\tan }^2}x}}{2}} \right) - 4\left( { - \ln \left| {\cos x} \right|} \right) + C \cr
& {\text{simplifying}} \cr
& = 2{\tan ^2}x - 4\ln \left| {\sec x} \right| + C \cr
& {\text{using logarithmic properties}} \cr
& = 2{\tan ^2}x - 2\ln \left( {{{\sec }^2}x} \right) + C \cr
& {\text{where }}{\sec ^2}x = 1 + {\tan ^2}x \cr
& = 2{\tan ^2}x - 2\ln \left( {1 + {{\tan }^2}x} \right) + C \cr} $$
