jmc

Expanding, we get $$\frac{(x+5)(x+2)}{x+1}=\frac{x^2+7x+10}{x+1}.$$By long division, $$\frac{x^2+7x+10}{x+1}=x+6+\frac{4}{x+1}=(x+1)+\frac{4}{x+1}+5.$$By AM-GM, $$(x+1)+\frac{4}{x+1}\ge 2\sqrt{(x+1)\cdot \frac{4}{x+1}}=4,$$so $(x+1)+\frac{4}{x+1}+5\ge 9.$

Equality occurs when $x=1,$ so the minimum value is $\boxed{9}.$