Answer
$\frac{1}{3}{\tan ^3}x + C$
Work Step by Step
$$\eqalign{
& \text{Let } I = \int {\left( {{{\tan }^2}x + {{\tan }^4}x} \right)} dx \cr
& {\text{Factor the integrand}} \cr
& I= \int {\left( {1 + {{\tan }^2}x} \right){{\tan }^2}x} dx \cr
& {\text{Rewrite the integrand }}{\tan ^2}x{\text{ using }}{\tan ^2}\theta = {\sec ^2}\theta - 1 \cr
& I = \int {\left( {1 + {{\sec }^2}x - 1} \right){{\tan }^2}x} dx \cr
& = \int {{{\sec }^2}x{{\tan }^2}x} dx \cr
& = \int {{{\tan }^2}x{{\sec }^2}x} dx \cr
& {\text{Let }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& {\text{Substituting}} \cr
& I = \int {{u^2}} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr
& I = \frac{1}{3}{u^3} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\tan x{\text{ for }}u \cr
& I = \frac{1}{3}{\tan ^3}x + C \cr} $$
