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Question Number 200299 by Calculusboy last updated on 16/Nov/23

Answered by witcher3 last updated on 17/Nov/23

$$\mathrm{x}\to \frac{1}{\mathrm{x}}$$$$\mathrm{\Omega }={\int }_{\mathrm{\infty }}^0-\frac{\frac{{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)}{\mathrm{x}}}{1+{\mathrm{x}}^2+{\mathrm{x}}^4}.\frac{{\mathrm{x}}^4}{{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}(\frac{\pi }{2}-{\tan }^{-1}\left(\mathrm{x}\right))}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}$$$$=-\mathrm{\Omega }+\frac{\pi }{4}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{dx}}^2}{(1+{\mathrm{x}}^2+{\mathrm{x}}^4)}=-\mathrm{\Omega }+\frac{\pi }{4}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}$$$${\iff 2\mathrm{\Omega }=\frac{\pi }{4}.\frac{2}{\sqrt{3}}{\tan }^{-1}(\frac{2\mathrm{x}}{\sqrt{3}}+\frac{1}{\sqrt{3}})]}_0^{\mathrm{\infty }}=\frac{\pi }{2\sqrt{3}}(\frac{\pi }{2}-\frac{\pi }{6})$$$$=\frac{{\pi }^2}{6\sqrt{3}}\Rightarrow \mathrm{\Omega }=\frac{{\pi }^2}{12\sqrt{3}}$$$$$$

Commented by Calculusboy last updated on 18/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by witcher3 last updated on 19/Nov/23

$$\mathrm{withe}\mathrm{Pleasur}$$

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