1. Let $x^2+\sqrt{2}x-8=0$ have roots $\alpha,\beta$.

By Vieta's formulas, $\alpha+\beta=-\sqrt{2},\qquad \alpha\beta=-8.$

2. We are given $U_n=\alpha^n+\beta^n.$ Since $\alpha,\beta$ satisfy the quadratic equation, each root $r$ satisfies $r^2+\sqrt{2}r-8=0 \implies r^2=-\sqrt{2}r+8.$ Multiplying by $r^{n-2}$, $r^n=-\sqrt{2}r^{n-1}+8r^{n-2}.$ Applying this for $r=\alpha,\beta$ and adding, $U_n=-\sqrt{2}U_{n-1}+8U_{n-2}.$ So, $U_{10}=-\sqrt{2}U_9+8U_8.$

3. Now compute the required expression: $\frac{U_{10}+\sqrt{2}U_9}{2U_8}.$ Using the recurrence result, $U_{10}+\sqrt{2}U_9=(-\sqrt{2}U_9+8U_8)+\sqrt{2}U_9=8U_8.$ Therefore, $\frac{U_{10}+\sqrt{2}U_9}{2U_8}=\frac{8U_8}{2U_8}=4.$

4. Hence the required integer is $\boxed{4}.$