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

log x

$\frac{d y}{d x}(x \log x)+y=2 \log x$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \Rightarrow P=\frac{1}{x \log x}, Q=\frac{2}{x}$ 

∴ I.F. = $e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x$

Hence (2) is the correct answer.