$\displaystyle (xy'-1)\ln x=2y$

$\displaystyle y'x\ln x -\ln x=2y$

$\displaystyle y'x\ln x -2y=\ln x$

$\displaystyle y'-\frac{2y}{x\ln x}=\frac{1}{x}$

so $\displaystyle P(x) = e^{\int x\ln x~dx}$

Help with that integration, please? Should I try substitution, or should I go with integration by parts?