Answer
$(-3,-1/2)\cup(2,\infty)$
Work Step by Step
1. Rewrite the inequality as $f(x)=\frac{x+2}{x+3}-\frac{x-1}{x-2}\lt0$
$f(x)=\frac{x^2-4-x^2-3x+x+3}{(x+3)(x-2)}=-\frac{2x+1}{(x+3)(x-2)}\lt0$
2. Cut points $-3,-1/2,2$, make a table as shown.
3. The solution is $(-3,-1/2)\cup(2,\infty)$
