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Question Number 67527 by mathmax by abdo last updated on 28/Aug/19

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1+x^2}{1+x^4}dx$$$$$$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

$$LetcalleditL$$$$L=2{\int }_0^{\mathrm{\infty }}\frac{1+x^2}{1+x^4}dxwhenchangingx=u^{\frac{1}{4}}\Rightarrow dx=\frac{u^{\frac{1}{4}-1}}{4}du$$$$L=2{\int }_0^{\mathrm{\infty }}\frac{1}{4}.\frac{u^{\frac{1}{4}-1}}{1+u}du+2{\int }_0^{\mathrm{\infty }}\frac{1}{4}.\frac{u^{\frac{1}{2}+\frac{1}{4}-1}}{1+u}du$$$$=\frac{\pi }{2sin\left(\frac{\pi }{4}\right)}+\frac{\pi }{2sin\left(\frac{3\pi }{4}\right)}=\pi \sqrt{2}$$

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