Definite integral of unspecified function

I stumbled upon this gem whilst doing tasks, and I can't seem to quite grasp the key to solving this. The answer should be $\dfrac{1}{2}$. While I have tried a few things, which resulted in $\dfrac{1}{2}$, I am quite unsure if my methods were legit. I have thought of substitution, but I get stuck half way. I would very much appreciate it if someone could help me solve this (practising for a test).

$$\int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}\text{d}x$$

Edit: Let $f$ be a positive continuous function. The task is simple if you could simply plug in a function, but it says not to. I think it means to solve this generally.

• What is your question? It will help if you post what you have tried and the community can see if your approach is valid. – Erik M Mar 7 '17 at 22:59
• If we know nothing about $f(x)$, there is nothing to solve. First of all, we even do not know if this integration is meaningful - e.g. is f(x)/(f(x) + f(1-x)) measurable? – Yujie Zha Mar 7 '17 at 23:09
• $f(1-x)$ is just $f(x)$ run backwards on the domain $[0,1]$, for whatever that may be worth. – Alfred Yerger Mar 7 '17 at 23:22
• I really do not know where to start. I have posted the whole task as it is supposed to be solved. This is the last place I can ask, my friends are all sleeping..If I could somehow substitute the core to make them equal, I could in theory get something like f(1/2) / 2f(1/2) which would be 1/2. – MCrypa Mar 7 '17 at 23:22
• Hint: $$I = \int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\frac{f(x)+f(1-x)-f(1-x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\left(1 - \frac{f(1-x)}{f(x)+f(1-x)}\right)dx=\;\;\cdots\;\;=1 - I$$ – dxiv Mar 7 '17 at 23:44

Notice that if you change the variable in the integration by $x = 1 - x$, you get:
$$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx= \int_1^0 \frac{f(1 - x)}{f(1 - x) + f(x)}d(1 - x) = \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx$$
And $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx + \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx = \int_0^1 \frac{f(x) + f(1 - x)}{f(x) + f(1 - x)}dx = 1$$ So $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx = \frac {1}{2}$$