Given that a function $f(x)$ satisfies $$f(x)+2 f\left(\frac{2 x-1}{x+1}\right)=\frac{x}{(x+1)\left(x^2+4\right)}$$ Then find $$\int_3^4 f\left(\frac{x-2}{2 x-1}\right) d x$$
My hard effort: Let $$h(x)=\frac{x}{(x+1)\left(x^2+4\right)}$$ The given functional equation is: $$f(x)+2 f\left(\frac{2 x-1}{x+1}\right)=h(x)$$ Replace $x$ with $\frac{2x-1}{x+1}$ above, then we get $$\boxed{f\left(\frac{2 x-1}{x+1}\right)+2 f\left(\frac{x-1}{x}\right)=h\left(\frac{2 x-1}{x+1}\right)}----(1)$$
Also from the functional equation, replace $x$ with $x+1$, we get $$\boxed{f(x+1)+2 f\left(\frac{2 x+1}{x+2}\right)=h(x+1)}----(2)$$
Now comes the interesting part: Let us assume $$I=\int_3^4 f\left(\frac{x-2}{2 x-1}\right) d x---(3)$$ Use the substitution $x=\frac{-1}{v}$, then we get $$\boxed{I=\int_{\frac{-1}{3}}^{\frac{-1}{4}} \frac{f\left(\frac{2 v+1}{v+2}\right)}{v^2} d v }----(4)$$ Also in the integral $(3)$, use the substitution $x=\frac{-1}{u-1}$, we get $$\boxed{I=\int_{\frac{2}{3}}^{\frac{3}{4}} \frac{f\left(\frac{2 u-1}{u+1}\right)}{(u-1)^2} d u}-----(5)$$ Now divide equation $(1)$ through out by $(x-1)^2$ and Integrate from the limits $\frac{2}{3} \to \frac{3}{4}$, we get $$\int_{\frac{2}{3}}^{\frac{3}{4}} \frac{f\left(\frac{2 x-1}{x+1}\right)}{(x-1)^2} d x+2 \int_{\frac{2}{3}}^{\frac{3}{4}} \frac{f\left(\frac{x-1}{x}\right)}{(x-1)^2} d x=\int_{\frac{2}{3}}^{\frac{3}{4}} \frac{h\left(\frac{2 x-1}{x+1}\right)}{(x-1)^2} d x$$ $$\implies I+2J=\int_{\frac{2}{3}}^{\frac{3}{4}} \frac{h\left(\frac{2 x-1}{x+1}\right)}{(x-1)^2} d x------(6)$$ Also divide equation $(2)$ through out by $x^2$ and integrate from limits $\frac{-1}{3}\to \frac{-1}{4}$,then we get $$\int_{\frac{-1}{3}}^{\frac{-1}{4}} \frac{f(x+1)}{x^2} d x+2 \int_{\frac{-1}{3}}^{\frac{-1}{4}} \frac{f\left(\frac{2 x+1}{x+2}\right)}{x^2} d x=\int_{\frac{-1}{3}}^{\frac{-1}{4}} \frac{h(x+1)}{x^2} d x$$ $$\Rightarrow K+2I=\int_{\frac{-1}{3}}^{\frac{-1}{4}} \frac{h(x+1)}{x^2} d x----(7)$$ Where $$J=\int_{\frac{2}{3}}^{\frac{3}{4}} \frac{f\left(\frac{x-1}{x}\right)}{(x-1)^2} d x,\:K=\int_{\frac{-1}{3}}^{\frac{-1}{4}} \frac{f(x+1)}{x^2} d x$$
Any inputs please?