Let $$\underset{0 \leqslant x \leqslant 2}{\text{Max}}\left\{\frac{9-x^2}{5-x}\right\} = \alpha$$ and $$\underset{0 \leqslant x \leqslant 2}{\text{Min}}\left\{\frac{9-x^2}{5-x}\right\} = \beta$$. If $$\int_{\beta - 8/3}^{2\alpha - 1} \text{Max}\left\{\frac{9-x^2}{5-x}, x\right\}dx = \alpha_1 + \alpha_2 \log_e\left(\frac{8}{15}\right)$$, then $$\alpha_1 + \alpha_2$$ is equal to ______

We need to find $$\alpha = \max_{0 \leq x \leq 2}\left\{\frac{9-x^2}{5-x}\right\}$$ and $$\beta = \min_{0 \leq x \leq 2}\left\{\frac{9-x^2}{5-x}\right\}$$.

Analyze $$h(x) = \frac{9-x^2}{5-x}$$.

$$h'(x) = \frac{-2x(5-x) + (9-x^2)}{(5-x)^2} = \frac{x^2 - 10x + 9}{(5-x)^2} = \frac{(x-1)(x-9)}{(5-x)^2}$$

On $$[0,2]$$: $$(x-9) < 0$$. So $$h'(x) > 0$$ for $$x < 1$$ and $$h'(x) < 0$$ for $$x > 1$$.

$$h(0) = \frac{9}{5}$$, $$h(1) = \frac{8}{4} = 2$$, $$h(2) = \frac{5}{3}$$.

$$\alpha = h(1) = 2$$, $$\beta = h(2) = \frac{5}{3}$$.

Set up the integral.

$$\beta - \frac{8}{3} = \frac{5}{3} - \frac{8}{3} = -1$$, $$2\alpha - 1 = 3$$.

$$\int_{-1}^{3} \max\left\{\frac{9-x^2}{5-x},\, x\right\}dx$$

Find where $$h(x) = x$$.

$$\frac{9-x^2}{5-x} = x \implies 9 - x^2 = 5x - x^2 \implies x = \frac{9}{5}$$

For $$x < \frac{9}{5}$$: $$h(0) = \frac{9}{5} > 0$$, so $$h(x) > x$$.

For $$x > \frac{9}{5}$$: $$h(3) = 0 < 3$$, so $$h(x) < x$$.

Evaluate the integral.

$$\int_{-1}^{9/5} h(x)\,dx + \int_{9/5}^{3} x\,dx$$

Second integral:

$$\int_{9/5}^{3} x\,dx = \left[\frac{x^2}{2}\right]_{9/5}^{3} = \frac{9}{2} - \frac{81}{50} = \frac{225 - 81}{50} = \frac{144}{50} = \frac{72}{25}$$

First integral: Using long division, $$\frac{9-x^2}{5-x} = \frac{x^2-9}{x-5} = x + 5 + \frac{16}{x-5}$$.

$$\int_{-1}^{9/5}\left(x + 5 + \frac{16}{x-5}\right)dx = \left[\frac{x^2}{2} + 5x + 16\ln|x-5|\right]_{-1}^{9/5}$$

At $$x = \frac{9}{5}$$: $$\frac{81}{50} + 9 + 16\ln\frac{16}{5}$$

At $$x = -1$$: $$\frac{1}{2} - 5 + 16\ln 6 = -\frac{9}{2} + 16\ln 6$$

Difference: $$\frac{81}{50} + 9 + \frac{9}{2} + 16\ln\frac{16}{5} - 16\ln 6$$

$$= \frac{81}{50} + \frac{27}{2} + 16\ln\frac{16}{30} = \frac{81 + 675}{50} + 16\ln\frac{8}{15}$$

$$= \frac{756}{50} + 16\ln\frac{8}{15} = \frac{378}{25} + 16\ln\frac{8}{15}$$

Combine and identify $$\alpha_1, \alpha_2$$.

Total $$= \frac{378}{25} + \frac{72}{25} + 16\ln\frac{8}{15} = \frac{450}{25} + 16\ln\frac{8}{15} = 18 + 16\log_e\frac{8}{15}$$

So $$\alpha_1 = 18$$ and $$\alpha_2 = 16$$.

$$\alpha_1 + \alpha_2 = 18 + 16 = 34$$

Answer: 34