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0-x-1-2-x-1-ln-2-x-1-dx-




Question Number 151503 by talminator2856791 last updated on 21/Aug/21
                  ∫_0 ^( ∞)  ((x−1)/( (√(2^x −1)) ln(2^x −1))) dx
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}\:\mathrm{ln}\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}\:{dx} \\ $$$$\: \\ $$
Answered by Kamel last updated on 21/Aug/21
  Ω=∫_0 ^(+∞) ((x−1)/( (√(2^x −1))Ln(2^x −1)))dx  t=2^x −1⇒x=((Ln(1+t))/(Ln(2)))  ∴ Ω=(1/(Ln^2 (2)))∫_0 ^(+∞) ((Ln(1+t)−Ln(2))/((1+t)(√t)Ln(t)))dt         =−(1/(Ln^2 (2)))∫_0 ^(+∞) ((Ln(1+u)−Ln(u)−Ln(2))/((1+u)(√u)Ln(u)))du  ∴ Ω=^((√u)=y) (1/(Ln^2 (2)))∫_0 ^(+∞) (dy/( (1+y^2 )))=(π/(2Ln^2 (2)))
$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{+\infty} \frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}{Ln}\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{dx} \\ $$$${t}=\mathrm{2}^{{x}} −\mathrm{1}\Rightarrow{x}=\frac{{Ln}\left(\mathrm{1}+{t}\right)}{{Ln}\left(\mathrm{2}\right)} \\ $$$$\therefore\:\Omega=\frac{\mathrm{1}}{{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{t}\right)−{Ln}\left(\mathrm{2}\right)}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}{Ln}\left({t}\right)}{dt} \\ $$$$\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{{Ln}\left(\mathrm{1}+{u}\right)−{Ln}\left({u}\right)−{Ln}\left(\mathrm{2}\right)}{\left(\mathrm{1}+{u}\right)\sqrt{{u}}{Ln}\left({u}\right)}{du} \\ $$$$\therefore\:\Omega\overset{\sqrt{{u}}={y}} {=}\frac{\mathrm{1}}{{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{{dy}}{\:\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{2}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)} \\ $$

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