problem: Find the smallest integer $x$ where the expression $\frac{\sqrt{x-2}}{x^2+x-6}$ is defined.
solution: For the numerator to be defined, the expression inside the square root must be non-negative. Therefore, we have  $$x-2\ge0.$$ Therefore, $x\ge2$. 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, $x\neq 2$, $x\neq -3$, and $x \ge2$. Therefore, the smallest integer value of $x$ for the expression to be defined is $\boxed{3}$.