Finding a function (?) and computing its definite integral So I've come across this exercise from one of my old highschool textbooks:
$$\text{If}\ 2f\bigg(\frac{x-2}{x+1}\bigg) +f\bigg(\frac{x+1}{x-2}\bigg) = x$$
Considering this, find :
$$\int_{\frac{1}{2}}^{\frac{3}{4}}f(x)dx$$
To be honest, I've been at it for quite a few minutes now and I can't really find a way to crack it and it's kinda embarassing.
I've done a little plot to help me guess the function, but that got me nowhere. I'm guessing we should use the fact that $\frac{x-2}{x+1}$ and $\frac{x+1}{x-2}$ are inverses, but I couldn't manage to solve for $f(x)$ .
Any ideas?
 A: So, for starters, you can show that $g(t) = \frac{-3}{t-1}-1$ is the inverse of $\frac{t-2}{t+1}$; i.e., composing one on the other gives $t$. Substituting $g(t)$ for $x$ thus gives $2f(t) + f(\frac{1}{t}) = g(t)$. Now observe that $2f(\frac{1}{t}) + f(t) = g(1/t).$ Thus $f(t) + f(\frac{1}{t}) = \frac{g(t) + g(\frac{1}{t})}{3}$. But then $$f(t) = \big[2f(t) + f(\frac{1}{t})\big] - \big[f(t) + f(\frac{1}{t})\big] = g(t) - \frac{g(t) + g(\frac{1}{t})}{3} $$
and we now know that $f(t)$ is a rational function of $t$, which is integrable. 
I'm assuming there's a more elegant way of doing this which elides over determining $f$ explicitly... but, this way works also.
A: Changing $y=\frac{x-2}{x+1}$ one has $x=\frac {y+2}{1-y}$ so the relation becomes $2f(y)+f(\frac1y)=\frac{y+2}{1-y}$.  Since $x\to\frac1x$ defines an involution we can get a linear system in $f(y)$ and $f(\frac1y)$: $$2f(y)+f(\frac1y)=\frac{y+2}{1-y}$$  $$2f(\frac 1y) + f(y)=\frac {1+2y}{y-1}$$ from which we obtain $$f(y)= \frac{4y+5}{3-3y}$$
Now $$\int_{\frac12}^{\frac34}\frac{4y+5}{3-3y}dy= \ln 8-\frac13$$
