Question: I found this problem on an old FB post and have just been humbled by it.

Let $f(x)$ be continuous on $[0,n]$ such that $n > 0$ and $f(x) + f(n - x)$ does not vanish on $[0,n]$ then evaluate

$$
\int\_0^n \frac{f(x)}{f(x) + f(n-x)}dx
$$

Any clues/hints would be appreciated.

Answer: 
Let $\displaystyle I = \int\_0^n \frac{f(x)}{f(x) + f(n-x)}dx$

Let $y = n-x$ Then $dy = -dx$ and
$\displaystyle I = -\int\_n^0 \frac{f(n-y)}{f(n-y) + f(y)}dy
= \int\_0^n \frac{f(n-x)}{f(n-x) + f(x)}dx$

So
\begin{align}
 2I &= I + I\\
 2I &= \int\_0^n \frac{f(x)}{f(x) + f(n-x)}dx 
 + \int\_0^n \frac{f(n-x)}{f(n-x) + f(x)}dx\\
 2I &= \int\_0^n \frac{f(x)+f(n-x)}{f(x) + f(n-x)}dx\\
 2I &= \int\_0^n dx\\
 2I &= n\\
 I &= \frac n2
\end{align}
Let $I$ be our integral. Make the substitution $u=n-x$. 

After you have completed the change of variable, change the dummy variable of integration back to $x$. We now have a new expression for $I$. Add the old expression, and *look*.