Question Number 152533 by mnjuly1970 last updated on 29/Aug/21
Answered by Kamel last updated on 29/Aug/21
$${I}\overset{{t}=\sqrt{{x}}} {=}\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}\right)−{Ln}\left({t}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$\:\:\:\:=−\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)−{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{2}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$$\:\:\:=\mathrm{4}\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$