Integral Function

$\begin{aligned} I & = \int_0^1 \frac {f(x)}{f(x)+f(1-x)}\ dx & \small \color{#3D99F6} \text{Using identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^1 \left(\frac {f(x)}{f(x)+f(1-x)} + \frac {f(1-x)}{f(1-x)+f(x)}\right) dx \\ & = \frac 12 \int_0^1 \frac {f(x)+f(1-x)}{f(x)+f(1-x)}\ dx \\ & = \frac 12 \int_0^1 dx \\ & = \boxed{\dfrac 12} \end{aligned}$

This problem should be easy cause we just need to calculat the integral of any $f(x)$ that continuous at $[0,1]$ , just take $f(x)=x$ the rest is trivial.

The hardest part is how to prove it, for any $f(x)$ that continuous at $[0,1]$