$\int_3^5 \frac{\sqrt{x} d x}{\sqrt{8-x}+\sqrt{x}}=$

Consider the integral $I = \int_3^5 \frac{\sqrt{x} \, dx}{\sqrt{8-x} + \sqrt{x}}$.

Using the substitution property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$ with $a=3$ and $b=5$, we obtain:

$I = \int_3^5 \frac{\sqrt{8-x} \, dx}{\sqrt{x} + \sqrt{8-x}}$

Adding both expressions for $I$ yields:

$2I = \int_3^5 \frac{\sqrt{x} + \sqrt{8-x}}{\sqrt{x} + \sqrt{8-x}} \, dx = \int_3^5 1 \, dx$

Evaluating the integral gives $2I = 2$, thus $I = 1$.

Final answer: $\boxed{1}$