Let $I = \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \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 $ Then $I$ equals
$ (a)\ \frac{1}{2} \\ (b)\ \frac{1}{3} \\ (c)\ \frac{1}{4} \\ (d)\ 1 $
I've tried to solve this and I also could do the whole integration down to the last variable but the process is tremendously lengthy and I have a term $log7$ remaining in my solution. I don't know where I've made a calculative mistake but I just want some hint to solve this problem in a easier and a compact way.
Also I've thought about change of variable transformations but couldn't think of any suitable one.
By symmetry we can see that $$\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 + x_3 - x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 - x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 -x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{-x_1 + x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4$$ So $$I=\frac 14 \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 + x_3 - x_4}{x_1 + x_2 + x_3 + x_4} + \frac{x_1 + x_2 - x_3 +x_4}{x_1 + x_2 + x_3 + x_4} + \frac{x_1 -x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} \frac{-x_1 + x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \frac 14 \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} 2 dx_1 dx_2 dx_3 dx_4=\frac 12.$$