Let Max 0 ⩽ x ⩽ 2 9 - x 2 5 - x = α and Min 0 ⩽ x ⩽ 2 9 - x 2 5 - x = &#946…

Here $y=\frac{9-x^2}{5-x}=5+x+\frac{16}{x-5}$

$\frac{dy}{dx}=1-\frac{16}{{\left(x-5\right)}^2}$

When $\frac{dy}{dx}=0\Rightarrow 1-\frac{16}{{\left(x-5\right)}^2}=0$

So critical point is $x=1$ in $\left[0,2\right]$

Now we get $y\left(0\right)=\frac{9}{5},y\left(1\right)=2,y\left(2\right)=\frac{5}{3}$

So $\alpha =2$ and $\beta =\frac{5}{3}$

$I={\int }_{-1}^3\max \left(\frac{9-x^2}{5-x},x\right)$ 

$I={\int }_{-1}^{\frac{9}{5}}\left(5+x+\frac{16}{x-5}\right)dx+{\int }_{\frac{9}{5}}^{3}xdx$

$={\left[5x+\frac{x^2}{2}+16\ln \left(x-5\right)\right]}_{-1}^{\frac{9}{5}}+{\left[\frac{x^2}{2}\right]}_{\frac{9}{5}}^3$

$I=18+16\ln \left(\frac{8}{15}\right)$

${\alpha }_1=18$ and ${\alpha }_2=16$

$\Rightarrow {\alpha }_1+{\alpha }_2=34$