Answer
$$\frac{{{{\tan }^3}u}}{3} - \tan u + u + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^4}u} du \cr
& = \int {{{\tan }^2}u} {\tan ^2}udu \cr
& {\text{use ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr
& = \int {{{\tan }^2}u} \left( {{{\sec }^2}u - 1} \right)du \cr
& = \int {\left( {{{\tan }^2}u{{\sec }^2}u - {{\tan }^2}u} \right)} du \cr
& = \int {{{\tan }^2}u{{\sec }^2}u} du - \int {{{\tan }^2}u} du \cr
& = \int {{{\tan }^2}u{{\sec }^2}u} du - \int {\left( {{{\sec }^2}u - 1} \right)} du \cr
& = \int {{{\tan }^2}u{{\sec }^2}u} du - \int {{{\sec }^2}u} du + \int {du} \cr
& {\text{find antiderivatives}} \cr
& = \frac{{{{\tan }^3}u}}{3} - \tan u + u + C \cr} $$
