The integrating factor of the differential equation $\dfrac{dy}{dx}\left(x\log _e\:x\right)+y=2\log _e\:x$ is given by

$\dfrac{dy}{dx}\left(x\log_{e}{x}\right)+y=2\log_{e}{x}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{1}{\left(x\log_{e}{x}\right)}y=\dfrac{2\log_{e}{x}}{\left(x\log_{e}{x}\right)}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{1}{\left(x\log_{e}{x}\right)}y=\dfrac 2x$ which is of the form $\dfrac{dy}{dx}+py=q$

Integrating factor$={e}^{\int{p}dx}$ where $p=\dfrac{1}{\left(x\log_{e}{x}\right)}$

Integrating factor$={e}^{\int{\dfrac{1}{\left(x\log_{e}{x}\right)}}dx}$

Take $t=\log_{e}{x}\Rightarrow dt=\dfrac{1}{x}dx$

$\int{\dfrac{1}{\left(x\log_{e}{x}\right)}}dx=\int{\dfrac{dt}{t}}=\ln{\left|t\right|}\ where\ t=\log_{e}{x}$

Integrating factor$={e}^{\int{p}dx}={e}^{\ln t}=t=\log_{e}{x}$