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Question Number 162792 by HongKing last updated on 01/Jan/22

$$\mathrm{Find}:$$$$\mathbf{\Omega }=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\arctan \left(\mathrm{x}\right)}{\mathrm{x}\cdot ({\mathrm{x}}^2-\mathrm{x}+1)}\mathrm{dx}$$

Answered by phanphuoc last updated on 01/Jan/22

$$wehavearctan\left(x\right)=arctan\left(tx\right){\mid }_0^1={\int }_0^1\frac{xdt}{1+{\left(tx\right)}^2}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{\int }_0^1\frac{dtdx}{(1+{\left(tx\right)}^2(1-x+x^2)}$$$$youcanfindf\left(t\right),g\left(t\right)$$$$1/\left(\left((1+{\left(tx\right)}^2)(1-x+x^2)\right)\right)=\frac{f\left(t\right)}{1+{\left(tx\right)}^2}+\frac{g\left(t\right)}{1-x+x^2}$$$$$$

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