Answer
$\frac{2(x+2)}{(x+1)} ; x \ne 1,-1,2,-2;$
Work Step by Step
$\frac{\frac{x}{x-2} + 1}{\frac{3}{x^{2}-4}+ 1}$
Simplifying the numerator,
$\frac{x}{x-2} + 1 = \frac{x + (x-2)}{(x-2)} = \frac{2x-2}{(x-2)} = \frac{2(x-1)}{(x-2)}$
Simplifying the denominator,
$\frac{3}{x^{2}-4}+ 1 = \frac{3+(x^{2}-4)}{(x^{2}-4)} = \frac{(x^{2}-1)}{(x^{2}-4)} = \frac{(x+1)(x-1)}{(x+2)(x-2)}$
[Using $ (a^{2}-b^{2}) = (a+b)(a-b) $]
Substituting numerator and denominator, the given expression becomes
$\frac{\frac{x}{x-2} + 1}{\frac{3}{x^{2}-4}+ 1} = \frac{\frac{2(x-1)}{(x-2)}}{\frac{(x+1)(x-1)}{(x+2)(x-2)}} ; x \ne 1,-1,2,-2;$
$ = \frac{2(x-1)}{(x-2)} \times \frac{(x+2)(x-2)}{(x+1)(x-1)} ; x \ne 1,-1,2,-2;$
Divide out common factors,
$= \frac{2(x+2)}{(x+1)} ; x \ne 1,-1,2,-2;$
