jmc

For the numerator to be defined, the expression inside the square root must be non-negative. Therefore, we have $$x-2\ge 0.$$ Therefore, $x\ge 2$. The expression is undefined when the denominator is equal to 0, so it is undefined when $$x^2+x-6=(x-2)(x+3)=0.$$ So for the expression to be defined, $x\ne 2$, $x\ne -3$, and $x\ge 2$. Therefore, the smallest integer value of $x$ for the expression to be defined is $\boxed{3}$.