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Question Number 137941 by Ñï= last updated on 08/Apr/21

               ∫_0 ^∞ ((x−1)/( (√(2^x −1))ln (2^x −1)))dx=(π/(2ln^2 2))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}{ln}\:\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{dx}=\frac{\pi}{\mathrm{2}{ln}^{\mathrm{2}} \mathrm{2}} \\ $$

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