jmc

We can factor the numerator to get $y=\frac{(x+1)(x+2)}{x+1}$. If we exclude the case where $x=-1$, the function is equivalent to $y=x+2$. However, because $x$ cannot equal $-1$, $y$ cannot equal 1. Therefore, the range is all real numbers except for 1, which we may write as $y\in \boxed{(-\infty ,1)\cup (1,\infty )}.$