Q. Let $\underset{0 \leq x \leq 2}{\text{Max}} \left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\underset{0 \leq x \leq 2}{\text{Min}} \left\{\frac{9-x^{2}}{5-x}\right\}=\beta$ If $\int\limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \text{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{ e }\left(\frac{8}{15}\right)$ then $\alpha_{1}+\alpha_{2}$ is equal to ____
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