The integrating factor of the differential equation $(x\log_ex)\frac{dy}{dx}+y=2\log_ex$ is

$\log_ex$

The correct answer is Option (1) → $\log_ex$

$ (x\log_{e}x)\frac{dy}{dx}+y=2\log_{e}x $

$\text{Rewrite in standard form: }$

$ \frac{dy}{dx}+\frac{1}{x\log_{e}x}\,y=\frac{2}{x} $

$\text{Here } P(x)=\frac{1}{x\log_{e}x}. $

$\text{Integrating factor } IF=e^{\int P(x)\,dx} =e^{\int \frac{1}{x\log_{e}x}\,dx } $

$ \int \frac{1}{x\log_{e}x}\,dx =\log(\log_{e}x) $

therefore $ IF=e^{\log(\log_{e}x)}=\log_{e}x $

The integrating factor is $\log_{e}x$.