Question Number 31967 by abdo imad last updated on 17/Mar/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctanx}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx}\:. \\ $$
Commented by abdo imad last updated on 19/Mar/18
![let put I =∫_0 ^∞ ((arctanx)/(x^2 +x+1))dx .ch.x=(1/t) give I = ∫_0 ^∞ (((π/2) −arctant)/((1/t^2 ) +(1/t) +1)) (dt/t^2 ) = ∫_0 ^∞ (((π/2) −arctant)/(1+t +t^2 )) dt =(π/2) ∫_0 ^∞ (dt/(t^(2 ) +t +1)) −I ⇒ 2I= (π/2) ∫_0 ^∞ (dt/(t^2 +t+1)) ⇒ I =(π/4) ∫_0 ^∞ (dt/(t^2 +t +1)) .but ∫_0 ^∞ (dt/(t^2 +t +1)) = ∫_0 ^∞ (dt/((t+(1/2))^2 +(3/4)))( .ch.t+(1/2) =((√3)/2) u) = ∫_(1/( (√3))) ^(+∞) (1/((3/4)(t^(2 ) +1))) ((√3)/2) du = (4/3) ((√3)/2) ∫_(1/( (√3))) ^(+∞) (du/(1+u^2 )) =((2(√3))/3) [arctan(u)]_((1/( (√3))) ) ^∞ =((2(√3))/3) ( (π/2) −arctan((1/( (√3))))) =((2(√3))/3) arctan((√3))=((2(√3))/3) (π/3) =((2π(√3))/9) ⇒ I =(π/4) ((2π(√3))/9) = ((π^2 (√3))/(18)) .](https://www.tinkutara.com/question/Q32073.png)
$${let}\:{put}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctanx}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx}\:\:.{ch}.{x}=\frac{\mathrm{1}}{{t}}\:\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\frac{\pi}{\mathrm{2}}\:−{arctant}}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{t}}\:+\mathrm{1}}\:\frac{{dt}}{{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\pi}{\mathrm{2}}\:−{arctant}}{\mathrm{1}+{t}\:+{t}^{\mathrm{2}} }\:{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}\:} \:+{t}\:+\mathrm{1}}\:−{I}\:\Rightarrow\:\mathrm{2}{I}=\:\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}}\:\:.{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\left(\:\:.{ch}.{t}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{u}\right) \\ $$$$=\:\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}\left({t}^{\mathrm{2}\:} \:+\mathrm{1}\right)}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{du}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{+\infty} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\left[{arctan}\left({u}\right)\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:} ^{\infty} \:=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\left(\:\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\sqrt{\mathrm{3}}\right)=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\frac{\pi}{\mathrm{3}}\:=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:=\:\frac{\pi^{\mathrm{2}} \:\sqrt{\mathrm{3}}}{\mathrm{18}}\:. \\ $$