Answer
$$\int_{-\pi/4}^{\pi/4}(x^3+x^4\tan x)dx=0$$
Work Step by Step
Denote the function under the integral $f(x)=x^3+x^4\tan x.$ Now we have $$f(-x)=(-x)^3+(-x)^4\tan(-x)=-x^3+x^4(-\tan x)=-(x^3+x^4\tan x)=-f(x)$$
i.e. the function $f$ is odd. Knowing that the integral of an odd function with the symmetric integration bounds (one bound is $-a$ and the other is $a$) is equal to $0$ we have that:
$$\int_{-\pi/4}^{\pi/4}(x^3+x^4\tan x)dx=0$$
