Answer
$\int_{-\pi/4}^{\pi/4}(x^3+x^4~tan~x)~dx = 0$
Work Step by Step
The function $~~f(x) = tan~x~~$ is an odd function while $~~g(x) = x^4~~$ is an even function. Then $~~x^4~tan~x~~$ is an odd function.
Also note that $~~x^3~~$ is an odd function.
Therefore:
$\int_{-\pi/4}^{\pi/4}(x^3+x^4~tan~x)~dx$
$= \int_{-\pi/4}^{\pi/4}x^3~dx +\int_{-\pi/4}^{\pi/4}x^4~tan~x~dx$
$ = 0+0$
$= 0$
