Answer
$\displaystyle \frac{-(x^{2}+2x-1)}{(x+1)(x-1)}, \qquad x\neq 1, -1$
Work Step by Step
Factor each denominator.
$x^{2}-1=(x+1)(x-1)\qquad $... (a difference of squares)
$LCD=(x+1)(x-1)$
$\displaystyle \frac{x+3}{x^{2}-1}-\frac{x+2}{x-1}=\frac{x+3}{(x+1)(x-1)}-\frac{x+2}{x-1}$
$=\displaystyle \frac{x+3}{(x+1)(x-1)}-\frac{x+2}{x-1}\times\frac{x+1}{x+1}$
$=\displaystyle \frac{x+3}{(x+1)(x-1)}-\frac{x^{2}+3x+2}{(x+1)(x-1)}$
$=\displaystyle \frac{x+3-x^{2}-3x-2}{(x+1)(x-1)}$
$=\displaystyle \frac{-x^{2}-2x+1}{(x+1)(x-1)}$
$=\displaystyle \frac{-(x^{2}+2x-1)}{(x+1)(x-1)}, \qquad x\neq 1, -1$
