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Question Number 127420 by liberty last updated on 29/Dec/20

$${\int }_0^{\mathrm{\infty }}\frac{x-1}{\sqrt{2^x-1}\ln (2^x-1)}dx?$$

Answered by mnjuly1970 last updated on 29/Dec/20

$$solution:$$$$t=\sqrt{2^x-1}\Rightarrow t^2+1=2^x$$$$\mathrm{\Omega }=\frac{1}{{ln}^2\left(2\right)}{\int }_0^{\mathrm{\infty }}\left(\frac{ln(t^2+1)-ln\left(2\right)}{2tln\left(t\right)}\right)\frac{2tdt}{(t^2+1)}$$$$=\frac{1}{{ln}^2\left(2\right)}{\int }_0^{\mathrm{\infty }}\frac{ln(t^2+1)-ln\left(2\right)}{ln\left(t\right)(t^2+1)}dt$$$$\mathrm{\Omega }\stackrel{t=\frac{1}{y}}{=}\frac{1}{{ln}^2\left(2\right)}{\int }_0^{\mathrm{\infty }}\frac{ln(y^2+1)-2ln\left(y\right)-ln\left(2\right)}{-ln\left(y\right)(1+y^2)}dy$$$$=-\mathrm{\Omega }+\frac{2}{{ln}^2\left(2\right)}{\int }_0^{\mathrm{\infty }}\frac{dy}{1+y^2}$$$$2\mathrm{\Omega }=\frac{2}{{ln}^2\left(2\right)}\left(\frac{\pi }{2}\right)\Rightarrow \mathrm{\Omega }=\frac{\pi }{2{ln}^2\left(2\right)}✓$$

Commented by liberty last updated on 30/Dec/20

sabiħ ... grazzi għat-tlestija tiegħek

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