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

Question

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 :/

Answer 1

The result is quite unimaginable:

According to https://www.wolframalpha.com/input/?i=int%282-x%2B%28x-1%29ln+x-%28lnx%29%5E2%29%2F%281%2Bxlnx%29%5E2,

$\int\dfrac{2-x+(x-1)\ln x-\ln^2x}{(1+x\ln x)^2}~dx=\dfrac{x-x\ln^2x+(x\ln x+1)\ln(x\ln x+1)}{x\ln x+1}+C$