Proof that $\int_{0}^{1} \frac{f(x)}{f(x) + f(1-x)}dx = 1/2$, where $f:[0,1]\rightarrow \mathbb{R}$ is strictly positive and continuous?

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I'm trying to prove this for Riemann integration (not very rigorously, just finding an outline for how the proof should go). The substitution $y=1-x$ seemed handy, as I got $$\int_0^1 \frac{f(x) - f(1-x)}{f(x) + f(1-x)}dx = 0,$$ but I could not continue after that. Can anybody give me some kind of hint?

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Then you know that$$\int_0^1\frac{f(x)}{f(x)+f(1-x)}\,\mathrm dx=\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}\,\mathrm dx.\tag1$$But it follows from the substitution $y=1-x$ that the RHS of $(1)$ is equal to its LHS. And since their sum is $1$, you're done.