The integrating factor of the differential equation :

\(\frac{dy}{dx}\) (x log x) + y = 2 log x

is : 

log x

\(\frac{dy}{dx}\) (x log x) + y = 2 log x

\(\frac{dy}{dx}\) + \(\frac{y}{(x log x)}\) = \(\frac{2}{x}\)

Compare with : y' + yP(x) = Q(x)

Integrating factor : e^{\(\int_{}^{}d\)x P}= e ^{\(\int_{}^{}d\)x 1/x log x} = e ^{log(log x)} = log x