Answer
$$\int(\tan^2 x+\tan^4 x)dx=\frac{\tan^3 x}{3}+C$$
Work Step by Step
$$A=\int(\tan^2 x+\tan^4 x)dx$$ $$A=\int\tan^2 x(1+\tan^2 x)dx$$ $$A=\int\tan^2 x\sec^2 xdx$$ ($1+\tan^2 x=\sec^2 x$)
Now let $u=\tan x$
Then $du=\sec^2 xdx$.
Therefore, $$A=\int u^2du$$ $$A=\frac{u^3}{3}+C$$ $$A=\frac{\tan^3 x}{3}+C$$
