Question Number 163854 by amin96 last updated on 11/Jan/22
$$\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{residu}}\:\boldsymbol{\mathrm{theorem}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}}\boldsymbol{\mathrm{dx}}=?\:\:\:\: \\ $$
Answered by Ar Brandon last updated on 11/Jan/22
![I=∫_0 ^∞ (x^2 /(x^4 +2x^2 +2))dx=(1/2)∫_0 ^∞ (((x^2 +(√2))+(x^2 −(√2)))/(x^4 +2x^2 +2))dx =(1/2)∫_0 ^∞ ((x^2 +(√2))/(x^4 +2x^2 +2))dx+(1/2)∫_0 ^∞ ((x^2 −(√2))/(x^4 +2x^2 +2))dx =(1/2)∫_0 ^∞ ((1+((√2)/x^2 ))/(x^2 +2+(2/x^2 )))dx+(1/2)∫_0 ^∞ ((1−((√2)/x^2 ))/(x^2 +2+(2/x^2 )))dx =(1/2)∫_0 ^∞ ((1+((√2)/x^2 ))/((x−((√2)/x))^2 +2+2(√2)))dx+(1/2)∫_0 ^∞ ((1−((√2)/x^2 ))/((x+((√2)/x))^2 +2−2(√2)))dx =(1/2)∫_(−∞) ^(+∞) (du/(u^2 +(2+2(√2))))+(1/2)∫_(+∞) ^(+∞) (dv/(v^2 +2−2(√2))) =(1/2)∙(1/( (√(2+2(√2)))))[arctan((u/( (√(2+2(√2))))))]_(−∞) ^(+∞) =(π/(2(√(2+2(√2)))))](Q163864.png)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)+\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\right)}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} }}{\left({x}−\frac{\sqrt{\mathrm{2}}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} }}{\left({x}+\frac{\sqrt{\mathrm{2}}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{du}}{{u}^{\mathrm{2}} +\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int_{+\infty} ^{+\infty} \frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}\left[\mathrm{arctan}\left(\frac{{u}}{\:\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}\right)\right]_{−\infty} ^{+\infty} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$
Commented by peter frank last updated on 11/Jan/22
$$\mathrm{great} \\ $$
Commented by amin96 last updated on 11/Jan/22
$${great}\:{sir}.\:{correct}\:{answer} \\ $$