The integrating factor of the differential equation $\frac{{}^{dy} }{dx}(x lo{g}_e x)+y=2lo{g}_{e}x$ is given by 

 Given the differential equation,
$\frac{dy}{dx}\left(x{\log }_{e}x\right)+y=2{\log }_{e}x\text{. }$
Divide both sides by $\left(x{\log }_{e}x\right)$, we have.
$\left(\frac{dy}{dx}\right)+\left(\frac{y}{x{\log }_{e}x}\right)=\left(\frac{2{\log }_{e}x}{x{\log }_{e}x}\right)$
$\therefore \frac{dy}{dx}+\left(\frac{1}{x{\log }_{e}x}\right)y=\left(\frac{2}{x}\right)$ which is linear in $y$.
lo, the integrating factor is:

$={\int }_e^{\frac{1}{x{\log }_{e^x}}dx}$
$$\begin{gathered} =e^{{\log }_e\left({\log }_{e}x\right)} \\ ={\log }_{e}x \end{gathered}$$