Intuitive explanation of integral identity I have been able to prove the identity
$$\int_{0}^{1} \frac{f(x)}{f(x)+f(1-x)} \, dx = \frac{1}{2}$$ for any continous $f:[0,1]\to[0,\infty)$ for which the integrand is defined,
with calculus, but I would like to know if there is an intuitive explanation of the identity.
 A: Hint There is symmetry at play. Make the change of variables $x\to 1-x$, then sum. The integrand, call it $g(x)$, has the property that $g(1-x)=1-g(x)$.
A: Consider
\begin{align}
I = \int_{0}^{1} \frac{f(x)}{f(x)-f(1-x)} \, dx 
\end{align}
and let $x = 1-t$ to obtain
\begin{align}
I &= \int_{1}^{0} \frac{f(1-t)}{f(1-t) - f(t)} \, (-1) dt \\
&= - \int_{0}^{1} \frac{f(1-t)}{f(t) - f(1-t)} \, dt.
\end{align}
Now adding the two integrals yields
\begin{align}
2 I &= \int_{0}^{1} \frac{f(x) - f(1-x)}{f(x) - f(1-x)} \, dx = \int_{0}^{1} \, dt = 1
\end{align}
and hence 
\begin{align}
\int_{0}^{1} \frac{f(x)}{f(x)-f(1-x)} \, dx = \frac{1}{2}. 
\end{align}
A: $$\int_0^1\frac{f(x)}{f(x)-f(1-x)}dx=\frac{1}{2}\int_0^1\frac{f(x)}{1-\frac{1}{2x}}\left(\frac{x-(1-x)}{f(x)-f(1-x)}\right)\frac{dx}{x}$$
My fuzzy intuition is that $\frac{x-(1-x)}{f(x)-f(1-x)}$ is a crude approximation to the the reciprocal of the derivative. Multiplying $f$ by an approximation to the reciprocal of the derivative gives an approximation of the $x$ value. That cancels with the $x$ in the denominator. So you are integrating the rational function $\frac{1}{1-\frac{1}{2x}}$ from $0$ to $1$. Normally, you would get $2\log(2)$, but the error of the approximations above is precisely enough to scale this down to $1$.
I get the sense that this is related to logarithmic differentiation, but I cannot make it precise.
A: It's because of mirror-image symmetry around 1/2.
There is a beautiful illustration that clarifies it, but this is my first posting and I don't know how to draw graphics in a post.
I'll try to explain in words:  For every point x on the left side of 1/2 (i.e. in the interval [0,1/2)) there is a corresponding point 1-x on the right side of 1/2 (i.e. in the interval (1/2,1]).
You'll notice that your integrand f(x)/(f(x)+f(1-x)) has the nice property that whatever value y it takes at x, it takes on the value 1-y at its corresponding mirror-image point 1-x.
(You can work this out easily enough by substituting (1-x) in for x and doing the algebra.)
So, we're taking the integral of the y values over the interval [0,1/2), and then we're taking the integral of 1-y (those same y values!) over the interval (1/2,1].  You can see that whatever the y values sum to on the left side gets canceled out by whatever their negatives sum to on the right side.
The only thing which remains is the 1 over the interval (1/2,1].  That results in an integral of 1*(1/2) = 1/2.
That's it.
A: The square $[0,1]\times [0,1]$ is a union of two sets $\{y \le g(x)\}$ and $\{y \ge g(x)\}$.
If $g$ has the property $g(1-x)=1-g(x)$ then the symmetry with respect to the center of the square takes one set into the other, hence they are congruent and  have the same area $1/2$.
(assume $f>0$ so $g(x) = \frac{f(x)}{f(x)+f(1-x)} \in (0,1)$) 
