How would you solve the following?

$$\int^1_0 \frac{\operatorname{Li}_2(-x)}{x} \log(1+x)\, \mathrm dx$$



$$\int \frac{\log(1+x)}{x} \, \mathrm dx=-\text{Li}_2(-x)$$

Added later

Using the trick given in the hint, then $$\int \frac{\operatorname{Li}_2(-x)}{x} \log(1+x)\, \mathrm dx=-\frac{\text{Li}_2(-x){}^2}{2}$$ and so $$\int^1_0 \frac{\operatorname{Li}_2(-x)}{x} \log(1+x)\, \mathrm dx=-\frac{\pi ^4}{288}$$

  • $\begingroup$ very nice hint there! $\endgroup$ – Chinny84 Sep 30 '14 at 10:58
  • $\begingroup$ @Chinny84. Thanks ! It is a so complex integral (at first glance) that there must be a trick ! $\endgroup$ – Claude Leibovici Sep 30 '14 at 11:08
  • $\begingroup$ No problem. I am committing all these nice tricks to my mathematical "toolbox"! So Thank you. $\endgroup$ – Chinny84 Sep 30 '14 at 12:15
  • 1
    $\begingroup$ @Chinny84. I really would enjoy to have a tour in your toolbox ! Cheers :-) $\endgroup$ – Claude Leibovici Sep 30 '14 at 16:27
  • $\begingroup$ Send me an email if you want to discuss anything offline. I will update my profile accordingly :). Keep well. $\endgroup$ – Chinny84 Sep 30 '14 at 21:58

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