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

ex log x log (log x) x

log x

$\frac{dy}{dx}+\frac{y}{x\,\log x}=\frac{2}{x};I.F.=e^{\int\frac{1}{x\,\log x}}dx$ put t = log x $⇒ dt = \frac{1}{x}dx=e^{\frac{1}{t}dt}=e^{ln\,t}=t$ ⇒ I.F. = log x