Question Number 171395 by pticantor last updated on 14/Jun/22
$$\int_{−\infty} ^{+\infty} \frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} }\boldsymbol{{dx}} \\ $$
Commented by infinityaction last updated on 19/Jun/22
$$\:\:\:{I}\:\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$\:\:\:{put}\:\:\:\:{t}\:=\:{x}^{\mathrm{4}} \\ $$$$\:{I}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\:\frac{\pi}{\mathrm{2sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}=\:\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$
Commented by infinityaction last updated on 14/Jun/22
![∫_(−∞) ^∞ (1/(x^2 + (1/x^2 )))dx = ∫_(−∞) ^∞ (dx/((x−(1/x))^2 + 2)) ∫_(−∞) ^∞ (dx/(x^2 +((√2))^2 )) = [((tan^(−1) (x/(√2)) )/( (√2)))]_(−∞) ^∞ (π/( (√2)))](https://www.tinkutara.com/question/Q171409.png)
$$\int_{−\infty} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx}\:=\:\int_{−\infty} ^{\infty} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\:\mathrm{2}} \\ $$$$\int_{−\infty} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:\left[\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}/\sqrt{\mathrm{2}}\right)\:}{\:\sqrt{\mathrm{2}}}\right]_{−\infty} ^{\infty} \\ $$$$\:\:\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$
Answered by floor(10²Eta[1]) last updated on 14/Jun/22
![∫_(−∞) ^∞ (x^2 /((x^2 +(√2)x+1)(x^2 −(√2)x+1)))dx ((1−i)/4)∫_(−∞) ^∞ (dx/(x^2 +(√2)x+1))+((1+i)/4)∫_(−∞) ^∞ (dx/(x^2 −(√2)x+1)) ((1−i)/4)∫_(−∞) ^∞ (dx/((x+((√2)/2))^2 +(1/2)))+((1+i)/4)∫_(−∞) ^∞ (dx/((x−((√2)/2))^2 +(1/2))) ((1−i)/4)[(√2)arctg((√2)x+1)]_(−∞) ^∞ +((1+i)/4)[(√2)arctg((√2)x−1)]_(−∞) ^∞ ((1−i)/4)(√2)((π/2)−(−(π/2)))+((1+i)/4)(√2)((π/2)−(−(π/2))) ((π(√2)−iπ(√2))/4)+((π(√2)+iπ(√2))/4)=((π(√2))/2)](https://www.tinkutara.com/question/Q171405.png)
$$\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$$$\frac{\mathrm{1}−\mathrm{i}}{\mathrm{4}}\int_{−\infty} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}+\frac{\mathrm{1}+\mathrm{i}}{\mathrm{4}}\int_{−\infty} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}−\mathrm{i}}{\mathrm{4}}\int_{−\infty} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}+\mathrm{i}}{\mathrm{4}}\int_{−\infty} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}−\mathrm{i}}{\mathrm{4}}\left[\sqrt{\mathrm{2}}\mathrm{arctg}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)\right]_{−\infty} ^{\infty} +\frac{\mathrm{1}+\mathrm{i}}{\mathrm{4}}\left[\sqrt{\mathrm{2}}\mathrm{arctg}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)\right]_{−\infty} ^{\infty} \\ $$$$\frac{\mathrm{1}−\mathrm{i}}{\mathrm{4}}\sqrt{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}+\mathrm{i}}{\mathrm{4}}\sqrt{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$$\frac{\pi\sqrt{\mathrm{2}}−\mathrm{i}\pi\sqrt{\mathrm{2}}}{\mathrm{4}}+\frac{\pi\sqrt{\mathrm{2}}+\mathrm{i}\pi\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$