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From the given inequality, $$\frac{x+1}{x+2}-\frac{3x+4}{2x+9}>0,$$which simplifies to $$-\frac{x^2-x-1}{(x+2)(2x+9)}>0,$$or $$\frac{x^2-x-1}{(x+2)(2x+9)}<0.$$The solutions to $x^2-x-1=0$ are $x=\frac{1\pm \sqrt{5}}{2}.$ We can fill in a sign chart as follows:

$$\begin{array}{cccccc}& x<-\frac{9}{2}& -\frac{9}{2}<x<-2& -2<x<\frac{1-\sqrt{5}}{2}& \frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}& \frac{1+\sqrt{5}}{2}<x\\ 2x+9& -& +& +& +& +\\ x+2& -& -& +& +& +\\ x-\frac{1-\sqrt{5}}{2}& -& -& -& +& +\\ x-\frac{1+\sqrt{5}}{2}& -& -& -& -& +\\ \frac{x^2-x-1}{(x+2)(2x+9)}& +& -& +& -& +\end{array}$$Thus, the solution to $\frac{x^2-x-1}{(x+2)(2x+9)}<0$ is $$x\in \boxed{\left(-\frac{9}{2},-2\right)\cup \left(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right)}.$$