Answer
$(-\infty,-3)\cup[-1,3)$
Work Step by Step
1. $\frac{x+1}{x^2-9}\le0 \Longrightarrow \frac{x+1}{(x+3)(x-3)}\le0$
2. Identify boundary points $x=-3,-1,3$ and form intervals $(-\infty,-3),(-3,-1],[-1,3),(3,\infty)$
3. Choose test values for each interval $x=-4,-2,0,4$ and test the inequality to get results $True,\ False,\ True,\ False$
4. Thus we have the solution $(-\infty,-3)\cup[-1,3)$
