Tough inequality problem, splitting into cases Find the set of all $x$ for which $\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x+1}$.
I am new to solving such problems and am not able to appropriately break up the problem into cases, so any help would be thoroughly appreciated.

$$\dfrac{2x}{2x^2+5x+2}>\dfrac{1}{x+1}\iff\frac{2x}{2x^2+5x+2}-\frac{1}{x+1}>0 \\ \iff\frac{2x(x+1)-(2x^2+5x+2)}{(2x^2+5x+2)(x+1)}>0 \\ \iff-\frac{3x+2}{(2x+1)(x+2)(x+1)}>0 \\ \iff \frac{3x+2}{(2x+1)(x+2)(x+1)}<0$$
We can see that the numerator vanishes when $x=-\frac{2}{3}$, also the fraction is discontinuous for $x\in\{-2,-1,-\frac{1}{2}\}$. Since rational functions are continuous in their domain we must analyze what happens for $x$ in each of the intervals $(-\infty,-2),\,(-2,-1),\,(-1,-2/3),\,(-2/3,-1/2)$ and $(-1/2,\infty)$.
