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