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calculate-1-x-2-1-x-4-dx-




Question Number 67527 by mathmax by abdo last updated on 28/Aug/19
calculate ∫_(−∞) ^(+∞)   ((1+x^2 )/(1+x^4 ))dx
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$ \\ $$
Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19
 Let  called it L  L=2∫_0 ^∞ ((  1+x^2 )/(1+x^4 )) dx      when changing   x=u^(1/4)     ⇒ dx=(u^((1/4)−1) /4)du  L=2∫_0 ^∞ (1/4).(u^((1/4)−1) /(1+u))du+2∫_0 ^∞   (1/4).(u^((1/2)+(1/4)−1) /(1+u))du     =(π/(2sin((π/4)))) +(π/(2sin(((3π)/4)))) =π(√2)
$$\:{Let}\:\:{called}\:{it}\:{L} \\ $$$${L}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\:\:\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:\:\:\:\:\:{when}\:{changing}\:\:\:{x}={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:\:\:\Rightarrow\:{dx}=\frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{4}}{du} \\ $$$${L}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{4}}.\frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{u}}{du}+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}.\frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{u}}{du}\: \\ $$$$\:\:=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:+\frac{\pi}{\mathrm{2}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\pi\sqrt{\mathrm{2}}\: \\ $$

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