Question Number 134962 by 0731619177 last updated on 09/Mar/21
Answered by Olaf last updated on 09/Mar/21
![∀x∈R^∗ , arctanx+arctan(1/x) = (π/2) Ω = ∫_0 ^∞ ((xarctanx)/(x^4 −x^2 +1)) dx Let u = (1/x) Ω = ∫_∞ ^0 (((1/u)arctan(1/u))/((1/u^4 )−(1/u^2 )+1)) (−(du/u^2 )) Ω = ∫_0 ^∞ ((u((π/2)−arctanu))/(u^4 −u^2 +1)) du ⇒ Ω = (π/4)∫_0 ^∞ (u/(u^4 −u^2 +1)) du ∫(u/(u^4 −u^2 +1)) du = (1/( (√3)))arctan[(1/( (√3)))(2x^2 −1)]+C ∫_0 ^∞ (u/(u^4 −u^2 +1)) du = (1/( (√3)))[(π/2)−arctan(−(1/( (√3))))] = (1/( (√3)))[(π/2)−(−(π/( 6)))] = ((2π)/( 3(√3))) Ω = (π/4)×((2π)/(3(√3))) = (π^2 /( 6(√3)))](https://www.tinkutara.com/question/Q134965.png)
$$\forall{x}\in\mathbb{R}^{\ast} ,\:\mathrm{arctan}{x}+\mathrm{arctan}\frac{\mathrm{1}}{{x}}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}\mathrm{arctan}{x}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$$\Omega\:=\:\int_{\infty} ^{\mathrm{0}} \frac{\frac{\mathrm{1}}{{u}}\mathrm{arctan}\frac{\mathrm{1}}{{u}}}{\frac{\mathrm{1}}{{u}^{\mathrm{4}} }−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\mathrm{1}}\:\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right) \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{u}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}{u}\right)}{{u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}}\:{du} \\ $$$$\Rightarrow\:\Omega\:=\:\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{u}}{{u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}}\:{du} \\ $$$$\int\frac{{u}}{{u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}}\:{du}\:= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{arctan}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)\right]+\mathrm{C} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{u}}{{u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}}\:{du}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right] \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\:\mathrm{6}}\right)\right]\:=\:\frac{\mathrm{2}\pi}{\:\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{4}}×\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\pi^{\mathrm{2}} }{\:\mathrm{6}\sqrt{\mathrm{3}}} \\ $$
Commented by 0731619177 last updated on 09/Mar/21
$${thank}\:{you}\:{sir} \\ $$
Commented by Olaf last updated on 09/Mar/21
$$\mathrm{sorry},\:\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{was}\:\mathrm{wrong}. \\ $$$$\mathrm{I}\:\mathrm{corrected}. \\ $$