Question Number 151503 by talminator2856791 last updated on 21/Aug/21
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}\:\mathrm{ln}\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}\:{dx} \\ $$$$\: \\ $$
Answered by Kamel last updated on 21/Aug/21
$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{+\infty} \frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}{Ln}\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{dx} \\ $$$${t}=\mathrm{2}^{{x}} −\mathrm{1}\Rightarrow{x}=\frac{{Ln}\left(\mathrm{1}+{t}\right)}{{Ln}\left(\mathrm{2}\right)} \\ $$$$\therefore\:\Omega=\frac{\mathrm{1}}{{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{t}\right)−{Ln}\left(\mathrm{2}\right)}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}{Ln}\left({t}\right)}{dt} \\ $$$$\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{u}\right)−{Ln}\left({u}\right)−{Ln}\left(\mathrm{2}\right)}{\left(\mathrm{1}+{u}\right)\sqrt{{u}}{Ln}\left({u}\right)}{du} \\ $$$$\therefore\:\Omega\overset{\sqrt{{u}}={y}} {=}\frac{\mathrm{1}}{{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{{dy}}{\:\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)} \\ $$