Verified answer

[tex]\displaystyle \frac{3(x+1)}{(x-2)(x+5)}\geq \frac{1}{x-1},x\neq 2,x\neq -5,x\neq 1\\ \\\frac{3(x+1)}{(x-2)(x+5)}-\frac{1}{x-1}\geq 0\\\\\frac{3(x-1)(x+1)-(x-2)(x+5)}{(x-2)(x+5)(x-1)}\geq 0\\ \\\frac{3x^2-3-(x^2+3x-10)}{(x-2)(x+5)(x-1)}\geq 0\\ \\\frac{2x^2+7-3x}{(x-2)(x+5)(x-1)}\geq 0\\ \\\left \{ {{2x^2+7-3x\geq 0} \atop {(x-2)(x+5)(x-1) > 0}} \right.\\\\ \left \{ {{2x^2+7-3x\leq 0} \atop {(x-2)(x+5)(x-1) < 0}} \right. \\\\x \in (-5,1) \cup (2,+ \infty)[/tex]