$\large\frac{dy}{dx}+ \frac{1}{x \log x }$$y =\large\frac{2}{x^2}$
$P= \large\frac{1}{x \log x }$
$\int Pdx =\int \large\frac{1}{\log x} . \frac{1}{x} $$dx$
$\qquad= \log (\log x)$
I.F $e^{\int Pdx}=e^{\log(\log x)} $
$\qquad=\log x$
Hence 2 is the correct answer.