Answer
$I=\frac{a}{2}$
Work Step by Step
Step 1. Letting $u=a-x$, we have $du=-dx$ and $x=0\to u=a$, $x=1\to u=0$
Step 2. Evaluate the integral
$I=\int_0^a\frac{f(x)}{f(x)+f(a-x)}dx=\int_a^0\frac{f(a-u)}{f(a-u)+f(u)}(-du)=\int_0^a\frac{f(a-u)}{f(a-u)+f(u)}du$
Step 3. Let
$J=\int_0^a\frac{f(a-u)}{f(a-u)+f(u)}du=\int_0^a\frac{f(a-x)}{f(a-x)+f(x)}dx$
(simply switching a symbol here), we have
$I+J=\int_0^a\frac{f(x)+f(a-x)}{f(a-x)+f(x)}dx=\int_0^a(1)dx=x|_0^a=a$
Step 4. Since $I=J$, we have $I=\frac{a}{2}$
