Integrating Factor

$\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF  $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$  

=  $e^{2x}\cdot {\left(x-1\right)}^2$  

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y =  $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$  

y(2) =  $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$  

y(3) =  $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$