Answer
$-\displaystyle \frac{2(x+2)}{x+1}\qquad x\neq 1,-1,2, -2$
Work Step by Step
Start by recognizing a difference of squares in the denominator of the denominator.
$ \displaystyle \frac{\frac{x}{x-2}+1}{\frac{3}{x^{2}-4}+1}= \frac{\frac{x}{x-2}+1}{\frac{3}{(x-2)(x+2)}-1}=...$
Complex rational expressions have rational expressions in the numerator or/and in the denominator.
Here, multiplying both the numerator and denominator with $(x-2)(x+2)$ would get rid of these fractions.
Currently, we note that we have to exclude $\pm 2$ as they yield zero in corresponding denominators..
The denominator of the complex rational expression must not be zero,
so we will exclude any values for which it is 0
$\displaystyle \frac{3}{(x-2)(x+2)}-1=0\quad/\times (x-2)(x+2)$
$3-1(x^{2}-4)=0$
$3-x^{2}+4=0$
$1-x^{2}=0$
$(1-x)(1+x)=0$
$x=\pm 1$ must be also excluded.
$\displaystyle \frac{\frac{x}{x-2}+1}{\frac{3}{(x-2)(x+2)}-1}\times\frac{(x-2)(x+2)}{(x-2)(x+2)}=\frac{x(x+2)+(x+2)(x-2)}{3-(x-2)(x+2)}$
... the denominator was simplified while solving the equation, above.
$=\displaystyle \frac{(x+2)[x+(x-2)]}{(1-x)(1+x)} $
$=\displaystyle \frac{(x+2)[2x-2]}{(1-x)(1+x)} $
$=\displaystyle \frac{(x+2)[2x-2]}{(1-x)(1+x)}\qquad$... factor out -2 from $2x-2$
$=-\displaystyle \frac{2(x+2)(1-x)}{(1-x)(x+1)}\qquad$... reduce the common factor
$=-\displaystyle \frac{2(x+2)}{x+1}\qquad x\neq 1,-1,2, -2$
