Answer
$\frac{x+3}{x+1}$
Work Step by Step
We multiply the numerator and denominator by $x+2$, then simplify:
$\displaystyle \frac{1+\frac{1}{x+2}}{1-\frac{1}{x+2}}=\frac{(x+2)(1+\frac{1}{x+2})}{(x+2)(1-\frac{1}{x+2})}=\frac{(x+2)+1}{(x+2)-1}=\frac{x+3}{x+1}$