If $f: [0,1]\rightarrow \mathbb{R}$ is continuous function positive If $f: [0,1]\rightarrow \mathbb{R}$ is continuous function positive, so $$\int_{0}^{1} \frac{f(x)}{f(x)+f(1-x)}dx=\frac{1}{2}$$???
all examples that I tested have worked.
 A: Hint:
$$
\int_{0}^{1} \left(
\frac{f(x)}{f(x)+f(1-x)}
+
\frac{f(1-x)}{f(x)+f(1-x)}
\right)
dx=1
$$

The idea is, after having noted the above fact, to split the sum and note that
$$
\int_{0}^{1}
\frac{f(1-x)}{f(x)+f(1-x)}\,dx=
\int_{1}^{0}
\frac{f(y)}{f(1-y)+f(y)}\,(-1)dy
$$
by doing the substitution $1-x=y$. Switching the integration bounds, and the variable back to $x$, we conclude that
$$
2\int_{0}^{1}
\frac{f(x)}{f(1-x)+f(x)}\,dx=1
$$
The assumption that $f$ is continuous doesn't by itself guarantee the integral exists, because we might have $f(x)+f(1-x)$, for some $x\in[0,1]$; but this can't happen if $f(x)>0$ for all $x\in[0,1]$.
A: Since
$$
\int_{1/2}^1\frac{f(x)}{f(x)+f(1-x)}\,dx\stackrel{y=1-x}{=}\int_0^{1/2}\frac{f(1-y)}{f(1-y)+f(y)}\,dy=\int_0^{1/2}\frac{f(1-x)}{f(x)+f(1-x)}\,dx,
$$
it follows that
\begin{eqnarray}
\int_0^1\frac{f(x)}{f(x)+f(1-x)}\,dx&=&\int_0^{1/2}\frac{f(x)}{f(x)+f(1-x)}\,dx+\int_{1/2}^1\frac{f(x)}{f(x)+f(1-x)}\,dx\\
&=&\int_0^{1/2}\frac{f(x)}{f(x)+f(1-x)}\,dx+\int_0^{1/2}\frac{f(1-x)}{f(x)+f(1-x)}\,dx\\
&=&\int_0^{1/2}\frac{f(x)+f(1-x)}{f(x)+f(1-x)}\,dx=\int_0^{1/2}\,dx=\frac12.
\end{eqnarray}
A: Forgot ?$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$
A: To egreg's answer I would add that
$$
\int_0^1
\frac{f(x)}{f(x)+f(1-x)} \,dx = 
\int_0^1
\frac{f(1-x)}{f(x)+f(1-x)}\,dx
$$
as can be seen via the substitution that says $u = 1-x$, $du = -dx$, and as $x$ goes from $0$ to $1$ then $u$ goes from $1$ to $0$.
