jmc

Let $y=\sqrt{x-1}.$ Then $y^2=x-1,$ so $x=y^2+1.$ Then $$\frac{x+8}{\sqrt{x-1}}=\frac{y^2+9}{y}=y+\frac{9}{y}.$$By AM-GM, $$y+\frac{9}{y}\ge 6.$$Equality occurs when $y=3,$ or $x=10,$ so the minimum value is $\boxed{6}.$