Let $\displaystyle \max_{0 \le x \le 2} \left\{ \frac{9-x^2}{5-x} \right\} = \alpha$ and $\displaystyle \min_{0 \le x \le 2} \left\{ \frac{9-x^2}{5-x} \right\} = \beta$.
If $\int_{\beta - \frac{8}{3}}^{2\alpha - 1} max \left\{ \frac{9-x^2}{5-x}, x \right\} dx = \alpha_1 + \alpha_2 \log_e \left( \frac{8}{15} \right)$ then $\alpha_1 + \alpha_2$ is equal to __________.
