Let $\operatorname{Max}_{0 \leq x \leq 2}\left\{\frac{9-x^2}{5-x}\right\}=\alpha$ and $\operatorname{Max}_{0 \leq x \leq 2}\left\{\frac{9-x^2}{5-x}\right\}=\beta$ If $\int_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9-x^2}{5-x}, x\right\} d x=\alpha_1+\alpha_2 \log _e\left(\frac{8}{15}\right)$ then $\alpha_1+\alpha_2$ is equal to $\_\_\_\_$