Answer
$\displaystyle \frac{-5x+4}{(x-2)(x+1)(x+3)}$
Work Step by Step
$\displaystyle \frac{\frac{x-2}{x+1}-\frac{x}{x-2}}{x+3}=(\frac{x-2}{x+1}-\frac{x}{x-2})\div(x+3)$
$=(\displaystyle \frac{x-2}{x+1}\cdot\frac{x-2}{x-2}-\frac{x}{x-2}\cdot\frac{x+1}{x+1})\div(x+3)$
$=(\displaystyle \frac{x^{2}-4x+4}{(x-2)(x+1)}-\frac{x^{2}-x}{(x-2)(x+1)})\div(x+3)$
$=(\displaystyle \frac{-5x+4}{(x-2)(x+1)})\div(x+3)$
... division = multiplication with the reciprocal,
$=\displaystyle \frac{-5x+4}{(x-2)(x+1)}\cdot\frac{1}{(x+3)}$
$=\displaystyle \frac{-5x+4}{(x-2)(x+1)(x+3)}$
