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Question Number 148564 by mathmax by abdo last updated on 29/Jul/21

calculate ∫_1 ^2 ((logx)/(1+x))dx

$$\mathrm{calculate}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{logx}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$

Answered by Kamel last updated on 29/Jul/21

Ω=∫_1 ^2 ((Ln(x))/(1+x))dx=−∫_(1/2) ^1 ((Ln(x))/(x(1+x)))dx =(1/2)Ln^2 (2)+Ln(2)Ln((3/2))−∫_(1/2) ^1 ((Ln(1+x))/x)dx =Ln(2)Ln(3)−(1/2)Ln^2 (2)+Li_2 (−1)−Li_2 (−(1/2)) ∴∫_1 ^2 ((Ln(x))/(1+x))dx =Ln(2)Ln(3)−(1/2)Ln^2 (2)−(π^2 /(12))−Li_2 (−(1/2))

$$ \\ $$$$\Omega=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{Ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}=−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Ln}\left({x}\right)}{{x}\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Ln}\left(\mathrm{2}\right){Ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$\:\:={Ln}\left(\mathrm{2}\right){Ln}\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{2}}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)−{Li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\therefore\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{Ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}\:={Ln}\left(\mathrm{2}\right){Ln}\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{2}}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−{Li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$

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