Answer
$$\frac{{{{\tan }^3}x}}{3} - \tan x + x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^4}x} dx \cr
& {\text{split exponent of }}{\tan ^4}x \cr
& = \int {{{\tan }^2}x} {\tan ^2}xdx \cr
& {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta \cr
& = \int {\left( {{{\sec }^2}x - 1} \right)} {\tan ^2}xdx \cr
& = \int {\left( {{{\sec }^2}x{{\tan }^2}x - {{\tan }^2}x} \right)} dx \cr
& = \int {\left( {{{\sec }^2}x{{\tan }^2}x - {{\sec }^2}x + 1} \right)} dx \cr
& {\text{sum rule}} \cr
& = \int {{{\tan }^2}x} {\sec ^2}xdx - \int {{{\sec }^2}x} dx + \int {dx} \cr
& {\text{find antiderivatives}} \cr
& = \frac{{{{\tan }^3}x}}{3} - \tan x + x + C \cr} $$
