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

First, put the equation in standard form $\frac{dy}{dx}+Py=Q$:
$$\frac{dy}{dx}+\frac{1}{x\log x}y=\frac{2}{x}$$
Here, $P=\frac{1}{x\log x}$.Integrating Factor (I.F.) = $e^{\int Pdx}$ $
$\text{I.F.}=e^{\int \frac{1}{x\log x}dx}$$Let$ u = \log x \implies du = \frac{1}{x} dx$.$$\int \frac{1}{u}du=\log u=\log (\log x)$ $
$$\text{I.F.}=e^{\log (\log x)}=\log x$$