Answer
$\displaystyle \frac{x^{2}-2x-5}{(x-2)(x+1)(x+2)}$
Work Step by Step
We factor the bottom terms and create a common denominator:
$\displaystyle \frac{1}{x+2}+\frac{1}{x^{2}-4}-\frac{2}{x^{2}-x-2}=\frac{1}{x+2}+\frac{1}{(x-2)(x+2)}-\frac{2}{(x-2)(x+1)}
=\frac{(x-2)(x+1)}{(x-2)(x+1)(x+2)}+\frac{x+1}{(x-2)(x+1)(x+2)}-\frac{2(x+2)}{(x-2)(x+1)(x+2)}$
We distribute and combine the top terms:
$=\displaystyle \frac{x^{2}-x-2+x+1-2x-4}{(x-2)(x+1)(x+2)}=\frac{x^{2}-2x-5}{(x-2)(x+1)(x+2)}$
