Answer
$\frac{-5}{(x-3)(x+2)(x+1)}$
Work Step by Step
We form a common denominator, then add, factor, and simplify:
$\displaystyle \frac{1}{x^{2}+3x+2}-\frac{1}{x^{2}-2x-3}=\frac{1}{(x+2)(x+1)}-\frac{1}{(x-3)(x+1)}
=\frac{x-3}{(x-3)(x+2)(x+1)}+\frac{-(x+2)}{(x-3)(x+2)(x+1)}=\frac{x-3-x-2}{(x-3)(x+2)(x+1)}=\frac{-5}{(x-3)(x+2)(x+1)}$
