The answer is an integer.

First we note that $x+2=\sqrt{5}$ and, cubing, $x^3+6x^2+12x+8=5\sqrt{5},$ or, $x^3+6x^2=5\sqrt{5}-8-12x$ . Now we can substitute $\frac{x^3+6x^2+13x+10}{x^3+6x^2+9x+2}=\frac{x+2+5\sqrt{5}}{-3x-6+5\sqrt{5}}=\frac{6\sqrt{5}}{2\sqrt{5}}=3$

$x = \sqrt{5} - 2$

$x+2 = \sqrt{5}$

$(x+2)^2 = (\sqrt{5})^2$

$x^2+4x+4 = 5$

$x^2 = 1-4x$

$x^3 = x-4x^2$

Simplify the given expression to $1+\frac{4x+8}{x^3+6x^2+9x+2} = 1+\frac{4x+8}{x-4x^2 + 6x^2 + 9x+2} =1+ \frac{4x+8}{2x^2+10x+2}$

$=1+\frac{4x+8}{2(1-4x) +10x+2} = 1+\frac{4x+8}{2-8x+10x+2} = 1+\frac{4x+8}{2x+4} = 1+2 = \boxed{3}$ .