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. 

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 $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx = \frac {1}{2}$$
