Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :/

share|cite|improve this question
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 '14 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 '14 at 17:40
How did @Paul commented with $1$ reputation? Was the rep requirement to comment finally abolished? :) – chubakueno Mar 30 '14 at 17:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.