$ \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_1dx_2dx_3dx_4 $ Evaluate
$$I=  \int_1^2\int_1^2 \int_1^2 \int_1^2  \frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4$$

Answer Options:


*

*$1$

*$\frac{1}{2}$

*$\frac{1}{3}$

*$\frac{1}{4}$


I need some suggestion here. I tried to evaluate one by one but it becomes messy. I think there some trick involved here.

An idea:
Since
$$\frac x{x+a}=1-\frac a{x+a}$$
we get that
$$\int_1^2\frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1=\int_1^2\left(1-\frac{x_2+x_3+x_4}{x_1+x_2+x_3+x_4}+\frac{x_2+x_3-x_4}{x_1+\ldots+x_4}\right)dx_1=$$
$$1-2x_4\int_1^2\frac{dx_1}{x_1+\ldots+x_4}=1-2x_4\log\frac{2+x_2+x_3+x_4}{1+x_2+x_3+x_4}$$
Try now with this the following iterations.
Another idea: Using Asal's very good comment:
Suppose
$$S=\int_1^2\int_1^2\int_1^2\int_1^2\frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4$$
As noted in the comment by Asal, the result is the same when we interchange the minus sign in the numerator among the different variables, because of symmetry, so if we call $\;A_i\;$ the multiple integral with the minus sign before $\;x_i\;$ in the numerator, we get
$$4S=A_1+A_2+A_3+A_4=2\int_1^2\int_1^2\int_1^2\int_1^2\frac{x_1+x_2+x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4=$$
$$=2\implies S=\frac12$$
