Algebra Questions

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

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

Answered by Kamel last updated on 01/Oct/21

$$\\$$ $$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({x}\right)β{Arctan}\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\frac{\mathrm{1}+{x}}{\mathrm{1}β{x}}{dx} \\$$ $$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left(\frac{{x}^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}{x}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\frac{\mathrm{1}+{x}}{\mathrm{1}β{x}}{dx};\:{t}=\frac{\mathrm{1}β{x}}{\mathrm{1}+{x}} \\$$ $$\:\:\:\:=β\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}=β\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}}{dt}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tArctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\$$ $$\:\:\:\:=β\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)β\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\$$ $$\:\:\:\Omega\:=β\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {Ln}\left({cos}\left({x}\right)\right){dx} \\$$ $$\:\:\:\:\:\:\:\:=β\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {Ln}\left({cos}\left({x}\right)\right){dx}β\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {Ln}\left({sin}\left({x}\right)\right){dx} \\$$ $$\:\:\:\:\:\:\:\:=β\mathrm{2}{G}+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)β\pi{Ln}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{4}}{Ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{xcos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\$$ $$\overset{{x}={Arctan}\left({t}\right)} {=}β\mathrm{2}{G}β\frac{\pi}{\mathrm{2}}{Ln}\left(\mathrm{2}\right)+\overset{=β\Omega} {\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}} \\$$ $$\therefore\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{Arctan}}\left(\boldsymbol{{x}}\right)β\boldsymbol{{Arctan}}\left(\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\:\frac{\mathrm{1}+\boldsymbol{{x}}}{\mathrm{1}β\boldsymbol{{x}}}\boldsymbol{{dx}}=β\boldsymbol{{G}}β\frac{\boldsymbol{\pi}}{\mathrm{4}}\boldsymbol{{Ln}}\left(\mathrm{2}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\$$

Commented bymathdanisur last updated on 02/Oct/21

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