jmc

We can write $x^2-6x+13=x^2-6x+9+4=(x-3{)}^2+4$. Therefore, since $(x-3{)}^2$ can never be negative, but we can make it zero when $x=3$, the smallest possible value of the expression $x^2-6x+13$ when $x$ is an integer is $\boxed{4}$.