Answer
$\frac{-x^{2}-2x+1}{(x+1)(x-1)}; x \ne 1,-1;$
Work Step by Step
$\frac{x+3}{x^{2}-1} - \frac{x+2}{x-1}$
$(x^{2}-1)$ is in the form of $(a^{2}-b^{2})$
$(a^{2}-b^{2}) = (a+b)(a-b)$
$(x^{2}-1)=(x^{2}-1^{2})= (x+1)(x-1)$
$= \frac{x+3}{(x+1)(x-1)} - \frac{x+2}{x-1}; x \ne 1,-1;$
Taking LCD,
$= \frac{(x+3)-(x+1)(x+2)}{(x+1)(x-1)} ; x \ne 1,-1;$
$= \frac{(x+3)-(x^{2}+3x+2)}{(x+1)(x-1)} ; x \ne 1,-1;$
$= \frac{x+3-x^{2}-3x-2}{(x+1)(x-1)} ; x \ne 1,-1;$
$=\frac{-x^{2}-2x+1}{(x+1)(x-1)}; x \ne 1,-1;$
