How to find $I=\int_{-4}^4\int_{-3}^3 \int_{-2}^2 \int_{-1}^1 \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4} \, dx_1 \, dx_2 \, dx_3 \, dx_4$

I couldn't resist as if your limits were same for each variable say varying $a$ to $b$, for any $a$ and $b$ , the following trick I saw somewhere, is applicable, but in your case, there is no short cut, I guess.

$$I=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{-x_1-x_2+x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{x_1-x_2-x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{-x_1+x_2{-}x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \int_{-a}^a \frac{x_1+x_2-x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{-x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4$$

Adding them all gives $6I=0 \implies I=0$

P.S- I know it doesn't answer your question, but it was too big/clumsy for a comment.