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Question Number 115689 by Algoritm last updated on 27/Sep/20

Answered by mathmax by abdo last updated on 27/Sep/20

$$\mathrm{I}={\int }_0^1\frac{\ln (\mathrm{x}+\sqrt{1-{\mathrm{x}}^2})}{\mathrm{x}}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{sint}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\ln (\mathrm{sint}+\mathrm{cost})}{\mathrm{sint}}\mathrm{cost}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{\ln (\mathrm{cost}+\mathrm{sint})}{\mathrm{sint}}.\mathrm{cost}\mathrm{dt}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\frac{\ln (\mathrm{cost}+\mathrm{asint})}{\mathrm{sint}}.\mathrm{cost}\mathrm{dt}\mathrm{with}\mathrm{a}>0$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cost}}{(\mathrm{cost}+\mathrm{asint})}\mathrm{dt}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}{\int }_0^1\frac{\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2}}{\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2}+\frac{2\mathrm{au}}{1+{\mathrm{u}}^2}}\times \frac{2\mathrm{du}}{1+{\mathrm{u}}^2}$$$$=2{\int }_0^1\frac{1-{\mathrm{u}}^2}{(1+{\mathrm{u}}^2)(1-{\mathrm{u}}^2+2\mathrm{au})}\mathrm{du}=2{\int }_0^1\frac{{\mathrm{u}}^2-1}{({\mathrm{u}}^2+1)({\mathrm{u}}^2-2\mathrm{au}-1)}\mathrm{du}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{u}\right)=\frac{{\mathrm{u}}^2-1}{({\mathrm{u}}^2+1)({\mathrm{u}}^2-2\mathrm{au}-1)}$$$${\mathrm{\Delta }}^{{}^{\prime }}={\mathrm{a}}^2+1>0\Rightarrow {\mathrm{u}}_1=\mathrm{a}+\sqrt{1+{\mathrm{a}}^2}\mathrm{and}{\mathrm{u}}_2=\mathrm{a}-\sqrt{1+{\mathrm{a}}^2}$$$$\Rightarrow \mathrm{F}\left(\mathrm{u}\right)=\frac{{\mathrm{u}}^2-1}{(\mathrm{u}-{\mathrm{u}}_1)(\mathrm{u}-{\mathrm{u}}_2)({\mathrm{u}}^2+1)}=\frac{\alpha }{\mathrm{u}-{\mathrm{u}}_1}+\frac{\beta }{\mathrm{u}-{\mathrm{u}}_2}+\frac{\mathrm{mu}+\mathrm{n}}{{\mathrm{u}}^2+1}$$$$\alpha =\frac{{\mathrm{u}}_1^2-1}{2\sqrt{1+{\mathrm{a}}^2}({\mathrm{u}}_1^2+1)}=\frac{{\mathrm{a}}^2+2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+1+{\mathrm{a}}^2-1}{2\sqrt{1+{\mathrm{a}}^2}({\mathrm{a}}^2+2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+1+{\mathrm{a}}^2+1)}$$$$=\frac{{2\mathrm{a}}^2+2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}}{2\sqrt{1+{\mathrm{a}}^2}({2\mathrm{a}}^2+2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+2)}=\frac{{\mathrm{a}}^2+\mathrm{a}\sqrt{1+{\mathrm{a}}^2}}{2\sqrt{1+{\mathrm{a}}^2}({\mathrm{a}}^2+\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+1)}$$$$\beta =\frac{{\mathrm{u}}_2^2-1}{-2\sqrt{1+{\mathrm{a}}^2}({\mathrm{u}}_2^2+1)}=\frac{{\mathrm{a}}^2-2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+1+{\mathrm{a}}^2-1}{-2\sqrt{1+{\mathrm{a}}^2}({\mathrm{a}}^2-2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+1+{\mathrm{a}}^2+1)}$$$$=\frac{{2\mathrm{a}}^2-2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}}{-2\sqrt{1+{\mathrm{a}}^2}({2\mathrm{a}}^2-2\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+2)}=\frac{\mathrm{a}\sqrt{1+{\mathrm{a}}^2}-{\mathrm{a}}^2}{2\sqrt{1+{\mathrm{a}}^2}({\mathrm{a}}^2-\mathrm{a}\sqrt{1+{\mathrm{a}}^2}+1)}$$$${\lim }_{\mathrm{u}\to +\mathrm{\infty }}\mathrm{uF}\left(\mathrm{u}\right)=0=\alpha +\beta +\mathrm{m}\Rightarrow \mathrm{m}=-\alpha -\beta$$$$\mathrm{F}\left(0\right)=\frac{-1}{(-1)}=1=-\frac{\alpha }{{\mathrm{u}}_1}-\frac{\beta }{{\mathrm{u}}_2}+\mathrm{n}\Rightarrow \mathrm{n}=1+\frac{\alpha }{{\mathrm{u}}_1}+\frac{\beta }{{\mathrm{u}}_2}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=2\alpha {\int }_0^1\frac{\mathrm{du}}{\mathrm{u}-{\mathrm{u}}_1}+2\beta {\int }_0^1\frac{\mathrm{du}}{\mathrm{u}-{\mathrm{u}}_2}+\mathrm{m}\ln ({\mathrm{u}}^2+1)+2\mathrm{n}\arctan \left(\mathrm{u}\right)$$$$=2\alpha \ln \mid \mathrm{u}-{\mathrm{u}}_1\mid +2\beta \ln \mid \mathrm{u}-{\mathrm{u}}_2\mid +\mathrm{mln}({\mathrm{u}}^2+1)+2\mathrm{narctan}\left(\mathrm{u}\right)+\mathrm{c}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=2\int \alpha \ln \mid \mathrm{u}-{\mathrm{u}}_1\mid \mathrm{da}+2\int \beta \ln \mid \mathrm{u}-{\mathrm{u}}_2\mid \mathrm{da}+\int \mathrm{mln}({\mathrm{u}}^2+1)\mathrm{da}$$$$+2\int \mathrm{narctan}\left(\mathrm{u}\right)\mathrm{da}+\mathrm{ca}+\lambda$$$$\alpha =\alpha \left(\mathrm{a}\right),\beta =\beta \left(\mathrm{a}\right),\mathrm{m}=\mathrm{m}\left(\mathrm{a}\right),\mathrm{n}=\mathrm{n}\left(\mathrm{a}\right)....\mathrm{be}\mathrm{continued}..\mathrm{a}\mathrm{lots}\mathrm{of}$$$$\mathrm{calculus}...$$

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