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Question Number 155512 by mathdanisur last updated on 01/Oct/21

𝛀 =∫_( 0) ^( 1) ((tan^(-1) x - tan^(-1) ((1/x)))/(1 + x^2 )) βˆ™ ((1+x)/(1-x)) dx = ?

$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{tan}^{-\mathrm{1}} \boldsymbol{\mathrm{x}}\:-\:\mathrm{tan}^{-\mathrm{1}} \left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\:\centerdot\:\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}-\mathrm{x}}\:\mathrm{dx}\:=\:? \\ $$

Answered by Kamel last updated on 01/Oct/21

Ξ©=∫_0 ^1 ((Arctan(x)βˆ’Arctan((1/x)))/(1+x^2 )) ((1+x)/(1βˆ’x))dx =∫_0 ^1 ((Arctan(((x^2 βˆ’1)/(2x))))/(1+x^2 )) ((1+x)/(1βˆ’x))dx; t=((1βˆ’x)/(1+x)) =βˆ’2∫_0 ^1 ((Arctan(t))/(t(1+t^2 )))dt=βˆ’2∫_0 ^1 ((Arctan(t))/t)dt+2∫_0 ^1 ((tArctan(t))/(1+t^2 ))dt =βˆ’2G+(Ο€/4)Ln(2)βˆ’βˆ«_0 ^1 ((Ln(1+t^2 ))/(1+t^2 ))dt Ξ© =βˆ’2G+(Ο€/4)Ln(2)+2∫_0 ^(Ο€/4) Ln(cos(x))dx =βˆ’2G+(Ο€/4)Ln(2)+2∫_0 ^(Ο€/2) Ln(cos(x))dxβˆ’2∫_0 ^(Ο€/4) Ln(sin(x))dx =βˆ’2G+(Ο€/4)Ln(2)βˆ’Ο€Ln(2)+(Ο€/4)Ln(2)+2∫_0 ^(Ο€/4) ((xcos(x))/(sin(x)))dx =^(x=Arctan(t)) βˆ’2Gβˆ’(Ο€/2)Ln(2)+2∫_0 ^1 ((Arctan(t))/(t(1+t^2 )))dt^(=βˆ’Ξ©) ∴ 𝛀=∫_0 ^1 ((Arctan(x)βˆ’Arctan((1/x)))/(1+x^2 )) ((1+x)/(1βˆ’x))dx=βˆ’Gβˆ’(𝛑/4)Ln(2) KAMEL BENAICHA

$$ \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({x}\right)βˆ’{Arctan}\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\frac{\mathrm{1}+{x}}{\mathrm{1}βˆ’{x}}{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left(\frac{{x}^{\mathrm{2}} βˆ’\mathrm{1}}{\mathrm{2}{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\frac{\mathrm{1}+{x}}{\mathrm{1}βˆ’{x}}{dx};\:{t}=\frac{\mathrm{1}βˆ’{x}}{\mathrm{1}+{x}} \\ $$$$\:\:\:\:=βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}=βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}}{dt}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tArctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:\:\:=βˆ’\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:\:\Omega\:=βˆ’\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {Ln}\left({cos}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:=βˆ’\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {Ln}\left({cos}\left({x}\right)\right){dx}βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {Ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:=βˆ’\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)βˆ’\pi{Ln}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{xcos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\overset{{x}={Arctan}\left({t}\right)} {=}βˆ’\mathrm{2}{G}βˆ’\frac{\pi}{\mathrm{2}}{Ln}\left(\mathrm{2}\right)+\overset{=βˆ’\Omega} {\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}} \\ $$$$\therefore\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{Arctan}}\left(\boldsymbol{{x}}\right)βˆ’\boldsymbol{{Arctan}}\left(\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\:\frac{\mathrm{1}+\boldsymbol{{x}}}{\mathrm{1}βˆ’\boldsymbol{{x}}}\boldsymbol{{dx}}=βˆ’\boldsymbol{{G}}βˆ’\frac{\boldsymbol{\pi}}{\mathrm{4}}\boldsymbol{{Ln}}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\ $$

Commented by mathdanisur last updated on 02/Oct/21

Very nice Ser thank you

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$

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