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Question Number 129001 by pipin last updated on 12/Jan/21

$$$$$${\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^4}=...$$

Answered by Ar Brandon last updated on 12/Jan/21

$$\mathrm{\Phi }={\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^4}=\frac{1}{2}{\int }_1^{\mathrm{\infty }}\{\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^4+1}-\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^4+1}\}\mathrm{dx}$$$$=\frac{1}{2}{\int }_1^{\mathrm{\infty }}\{\frac{1+\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}}-\frac{1-\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}}\}\mathrm{dx}=\frac{1}{2}{\int }_1^{\mathrm{\infty }}\{\frac{1+\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+2}-\frac{1-\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-2}\}\mathrm{dx}$$$$=\frac{1}{2}\{{\int }_0^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+2}-{\int }_2^{\mathrm{\infty }}\frac{\mathrm{dv}}{{\mathrm{v}}^2-2}\}=\frac{1}{2}\{{\left[\frac{{\tan }^{-1}\left(\mathrm{u}/\sqrt{2}\right)}{\sqrt{2}}\right]}_0^{\mathrm{\infty }}-\frac{1}{2\sqrt{2}}{[\ln \mid \frac{\sqrt{2}+\mathrm{v}}{\sqrt{2}-\mathrm{v}}\mid ]}_2^{\mathrm{\infty }}\}$$$$=\frac{1}{2}\{\frac{\pi }{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\ln \mid \frac{\sqrt{2}+2}{\sqrt{2}-2}\mid \}$$

Commented by pipin last updated on 12/Jan/21

$$\mathrm{omg},\mathrm{thankyou}\mathrm{bro}$$

Commented by Ar Brandon last updated on 12/Jan/21

You're welcome bro.

Answered by bramlexs22 last updated on 12/Jan/21

$${\int }_1^{\mathrm{\infty }}\frac{\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}={\int }_1^{\mathrm{\infty }}\frac{\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-2}\mathrm{dx}$$$${\int }_1^{\mathrm{\infty }}\frac{\frac{1}{{\mathrm{x}}^2}}{(\mathrm{x}+\frac{1}{\mathrm{x}}+\sqrt{2})(\mathrm{x}+\frac{1}{\mathrm{x}}-\sqrt{2})}\mathrm{dx}$$$$\mathrm{let}\frac{1}{\mathrm{x}}=\mathrm{u}\to \{\begin{array}{l}\mathrm{x}=1\to \mathrm{u}=1\\ \mathrm{x}=\mathrm{\infty }\to \mathrm{u}=0\end{array}$$$${\int }_1^0\frac{-\mathrm{du}}{(\mathrm{u}+\frac{1}{\mathrm{u}}+\sqrt{2})(\mathrm{u}+\frac{1}{\mathrm{u}}-\sqrt{2})}=$$$${\int }_0^1\frac{\mathrm{du}}{(\mathrm{u}+\frac{1}{\mathrm{u}}+\sqrt{2})(\mathrm{u}+\frac{1}{\mathrm{u}}-\sqrt{2})}$$

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