Let $y=y(x), x>1$, be the solution of the differential equation $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$, with $y(2)=\frac{1+e}{2 e^4}$. If $y(3) =\frac{\mathrm{e}^\alpha+1}{\beta \mathrm{e}^\alpha}$. Then the value of $\alpha+\beta$ is equal to $\_\_\_\_$ .
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