For any given function $f\colon [0,1]\to\Bbb R$, what is $\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx$?

Hint. Set $\displaystyle I=\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx$. By the change of variable $x \to1-x$ you get that $$ I=\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx. $$ Then observe that $$ I+I=\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx+\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx=\int_0^1\frac{f(x)+f(1-x)}{f(x)+f(1-x)}dx=1 $$ giving easily $$I=\frac12.$$