Let y = y(x), x > 1, be the solution of the differential equation $(x - 1){{dy} \over {dx}} + 2xy = {1 \over {x - 1}}$, with $y(2) = {{1 + {e^4}} \over {2{e^4}}}$. If $y(3) = {{{e^\alpha } + 1} \over {\beta {e^\alpha }}}$, then the value of $\alpha + \beta$ is equal to _________.

$\frac{d y}{d x}+y\left(\frac{2 x}{x-1}\right)=\frac{1}{(x-1)^{2}}$ <br/><br/> $\text { I.F} =e^{\int \frac{2 x}{x-1} d x}$ <br/><br/> $$ \begin{aligned} & =e^{2 \int\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right) d x} \\\\ & =e^{2 x+2 \ln (x-1)} \\\\ & =\mathrm{e}^{2 x}(\mathrm{x}-1)^{2} \end{aligned} $$ <br/><br/> $\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x$ <br/><br/> $\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c$ <br/><br/> $\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}$ <br/><br/> $\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c$ <br/><br/> $\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}$ <br/><br/> $\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}$ <br/><br/> $\downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}$ <br/><br/> $\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}$ <br/><br/> $\Rightarrow \alpha=6$ and $\beta=8 \Rightarrow \alpha+\beta=14$