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Question Number 155512 by mathdanisur last updated on 01/Oct/21

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

Answered by Kamel last updated on 01/Oct/21

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{Arctan\left(x\right)-Arctan\left(\frac{1}{x}\right)}{1+x^2}\frac{1+x}{1-x}dx$$$$={\int }_0^1\frac{Arctan\left(\frac{x^2-1}{2x}\right)}{1+x^2}\frac{1+x}{1-x}dx;t=\frac{1-x}{1+x}$$$$=-2{\int }_0^1\frac{Arctan\left(t\right)}{t(1+t^2)}dt=-2{\int }_0^1\frac{Arctan\left(t\right)}{t}dt+2{\int }_0^1\frac{tArctan\left(t\right)}{1+t^2}dt$$$$=-2G+\frac{\pi }{4}Ln\left(2\right)-{\int }_0^1\frac{Ln(1+t^2)}{1+t^2}dt$$$$\mathrm{\Omega }=-2G+\frac{\pi }{4}Ln\left(2\right)+2{\int }_0^{\frac{\pi }{4}}Ln\left(cos\left(x\right)\right)dx$$$$=-2G+\frac{\pi }{4}Ln\left(2\right)+2{\int }_0^{\frac{\pi }{2}}Ln\left(cos\left(x\right)\right)dx-2{\int }_0^{\frac{\pi }{4}}Ln\left(sin\left(x\right)\right)dx$$$$=-2G+\frac{\pi }{4}Ln\left(2\right)-\pi Ln\left(2\right)+\frac{\pi }{4}Ln\left(2\right)+2{\int }_0^{\frac{\pi }{4}}\frac{xcos\left(x\right)}{sin\left(x\right)}dx$$$$\stackrel{x=Arctan\left(t\right)}{=}-2G-\frac{\pi }{2}Ln\left(2\right)+\stackrel{=-\mathrm{\Omega }}{2{\int }_0^1\frac{Arctan\left(t\right)}{t(1+t^2)}dt}$$$$\therefore \mathbf{\Omega }={\int }_0^1\frac{\mathit{A}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)-\mathit{A}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\frac{1}{\mathit{x}}\right)}{1+{\mathit{x}}^2}\frac{1+\mathit{x}}{1-\mathit{x}}\mathit{d}\mathit{x}=-\mathit{G}-\frac{\mathit{\pi }}{4}\mathit{L}\mathit{n}\left(2\right)$$$$\mathit{K}\mathit{A}\mathit{M}\mathit{E}\mathit{L}\mathit{B}\mathit{E}\mathit{N}\mathit{A}\mathit{I}\mathit{C}\mathit{H}\mathit{A}$$

Commented by mathdanisur last updated on 02/Oct/21

$$\mathrm{Very}\mathrm{nice}\mathrm{Ser}\mathrm{thank}\mathrm{you}$$

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