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0-1-ln-1-x-ln-1-x-dx-by-MATH-AMIN-

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Question Number 163751 by amin96 last updated on 10/Jan/22
∫_0 ^1 ln(1+x)ln(1−x)dx=? by MATH.AMIN −−−−−−−−−−−−−−−−−−−−
$$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{x}}\right)\boldsymbol{{dx}}=? \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{MATH}}.\boldsymbol{{AMIN}} \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$
Answered by Kamel last updated on 10/Jan/22
Ω=−(1/2)(∫_0 ^1 Ln^2 (((1−x)/(1+x)))dx−∫_0 ^1 Ln^2 (1−x)dx−∫_0 ^1 Ln^2 (1+x)dx) =−(1/2)(2∫_0 ^1 ((Ln^2 (t))/((1+t)^2 ))dt−2−2Ln^2 (2)+2∫_0 ^1 Ln(1+x)dx) =−(1/2)(−4∫_0 ^1 ((Ln(t))/(1+t))dg−2−2Ln^2 (2)+4Ln(2)−2) =2+Ln^2 (2)−2Ln(2)−(π^2 /6)
$$\Omega=−\frac{\mathrm{1}}{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right){dx}\right) \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}^{\mathrm{2}} \left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}−\mathrm{2}−\mathrm{2}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {Ln}\left(\mathrm{1}+{x}\right){dx}\right) \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left({t}\right)}{\mathrm{1}+{t}}{dg}−\mathrm{2}−\mathrm{2}{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{4}{Ln}\left(\mathrm{2}\right)−\mathrm{2}\right) \\ $$$$\:\:=\mathrm{2}+{Ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{2}{Ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

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