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Question Number 33983 by abdo imad last updated on 28/Apr/18

$$find{\int }_{1(}^{\mathrm{\infty }}\frac{1}{x}ln\left(\frac{x+1}{x-1}\right)dx.$$

Commented by abdo mathsup 649 cc last updated on 03/May/18

$$letputI={\int }_1^{+\mathrm{\infty }}\frac{1}{x}ln\left(\frac{x+1}{x-1}\right)dxchangement$$$$x=\frac{1}{t}giveI={\int }_0^{1}tln\left(\frac{1+t}{1-t}\right)\frac{dt}{t^2}$$$$={\int }_0^1\frac{1}{t}ln\left(\frac{1+t}{1-t}\right)dt=\frac{{\pi }^2}{4}thisresultisproved.$$

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