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Question Number 134962 by 0731619177 last updated on 09/Mar/21

Answered by Olaf last updated on 09/Mar/21

$$\mathrm{\forall }x\in {\mathbb{R}}^{\ast },\arctan x+\arctan \frac{1}{x}=\frac{\pi }{2}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x\arctan x}{x^4-x^2+1}dx$$$$\mathrm{Let}u=\frac{1}{x}$$$$\mathrm{\Omega }={\int }_{\mathrm{\infty }}^0\frac{\frac{1}{u}\arctan \frac{1}{u}}{\frac{1}{u^4}-\frac{1}{u^2}+1}(-\frac{du}{u^2})$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{u(\frac{\pi }{2}-\arctan u)}{u^4-u^2+1}du$$$$\Rightarrow \mathrm{\Omega }=\frac{\pi }{4}{\int }_0^{\mathrm{\infty }}\frac{u}{u^4-u^2+1}du$$$$\int \frac{u}{u^4-u^2+1}du=$$$$\frac{1}{\sqrt{3}}\arctan \left[\frac{1}{\sqrt{3}}(2x^2-1)\right]+\mathrm{C}$$$${\int }_0^{\mathrm{\infty }}\frac{u}{u^4-u^2+1}du=\frac{1}{\sqrt{3}}[\frac{\pi }{2}-\arctan (-\frac{1}{\sqrt{3}})]$$$$=\frac{1}{\sqrt{3}}[\frac{\pi }{2}-(-\frac{\pi }{6})]=\frac{2\pi }{3\sqrt{3}}$$$$\mathrm{\Omega }=\frac{\pi }{4}\times \frac{2\pi }{3\sqrt{3}}=\frac{{\pi }^2}{6\sqrt{3}}$$

Commented by 0731619177 last updated on 09/Mar/21

$$thankyousir$$

Commented by Olaf last updated on 09/Mar/21

$$\mathrm{sorry},\mathrm{I}\mathrm{think}\mathrm{I}\mathrm{was}\mathrm{wrong}.$$$$\mathrm{I}\mathrm{corrected}.$$

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