# How to evaluate the integral : $I=\int\frac{2-x+(x-1)\ln x-\ln^2x}{(1+x\ln x)^2}dx$

How to evaluate the integral: $$I=\int\frac{2-x+(x-1)\ln x-\ln^2x}{(1+x \ln x)^2}dx.$$

Help me, thanks :/

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One observation is that the numerator factors as $(1-\ln x)(2-x+\ln x)$; according to Mathematica, $\frac{2-x+\ln x}{(1+x\ln x)^2}$ is integrable, and integration by parts works. But I don't immediately see how to integrate $\frac{2-x+\ln x}{(1+x\ln x)^2}$ nor the other resulting integral. –  rogerl Mar 30 at 17:35
Looks a bit like the derivative of u/v where v = 1 + xln(x). Play with that a bit to find a suitable u. Paul –  Paul Mar 30 at 17:40
How did @Paul commented with $1$ reputation? Was the rep requirement to comment finally abolished? :) –  chubakueno Mar 30 at 17:55