Answer
$-\displaystyle \frac{(x+3)(x-1)}{(x+2)(x-2)},\qquad x\neq 2, -2$
Work Step by Step
Factor each denominator.
$x^{2}-4=(x+2)(x-2)\qquad $... (a difference of squares)
$LCD=(x+2)(x-2)$
$\displaystyle \frac{x+5}{x^{2}-4}-\frac{x+1}{x-2}=\frac{x+5}{(x+2)(x-2)}-\frac{x+1}{x-2}$
$=\displaystyle \frac{x+5}{(x+2)(x-2)}-\frac{x+1}{x-2}\times\frac{x+2}{x+2}$
$=\displaystyle \frac{x+5}{(x+2)(x-2)}-\frac{x^{2}+3x+2}{(x+2)(x-2)}$
$=\displaystyle \frac{x+5-x^{2}-3x-2}{(x+2)(x-2)}$
$=\displaystyle \frac{-x^{2}-2x+3}{(x+2)(x-2)}$
$=\displaystyle \frac{-(x^{2}+2x-3)}{(x+2)(x-2)}$
... two factors of -3 with sum +2 are $+3$ and $-1$
$=-\displaystyle \frac{(x+3)(x-1)}{(x+2)(x-2)},\qquad x\neq 2, -2$
