jmc

Subtracting 1 from both sides and putting everything over a common denominator, we get $$\frac{-x^2+x+6}{(x+1)(x+5)}\ge 0.$$Equivalently, $$\frac{x^2-x-6}{(x+1)(x+5)}\le 0.$$We can factor the numerator, to get $$\frac{(x-3)(x+2)}{(x+1)(x+5)}\le 0.$$We build a sign chart, accordingly. \begin{array}{c|cccc|c} &$x-3$ &$x+2$ &$x+1$ &$x+5$ &$f(x)$ \\ \hline$x<-5$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$-5<x<-2$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$-2<x<-1$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$-1<x<3$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>3$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Also, note that $\frac{(x-3)(x+2)}{(x+1)(x+5)}=0$ for $x=-2$ and $x=3.$ Therefore, the solution is $$x\in \boxed{(-5,-2]\cup (-1,3]}.$$