Answer
$\frac{(x-3)}{(x+2)}; x \ne 3,-1,-2;$
Work Step by Step
$\frac{\frac{1}{x+1}}{\frac{1}{x^{2}-2x-3} + \frac{1}{x-3}}$
Factors of $x^{2}-2x-3$ are $(x-3),(x+1)$
$=\frac{\frac{1}{x+1}}{\frac{1}{(x-3)(x+1)} + \frac{1}{(x-3)}}$
Taking LCD in the Denominator,
$=\frac{\frac{1}{x+1}}{\frac{1+x+1}{(x-3)(x+1)}}$
$=\frac{\frac{1}{x+1}}{\frac{x+2}{(x-3)(x+1)}}; x \ne 3,-1,-2;$
$=\frac{1}{x+1} \times \frac{(x-3)(x+1)}{x+2}; x \ne 3,-1,-2;$
$= \frac{(x-3)}{(x+2)}; x \ne 3,-1,-2;$
