jmc

We begin by factoring the denominator: $y=\frac{x+1}{(x-1{)}^2}$. There is a vertical asymptote at $x=a$ for a rational function if the denominator is zero when $x=a$ (except when $x-a$ is also a factor of the numerator and has the same multiplicity as it does in the denominator). The only value of $x$ where that occurs is $x=\boxed{1}$.