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Question Number 132191 by rs4089 last updated on 12/Feb/21

$${\int }_0^1\frac{{log}_e(x+1)}{x(x^2+1)}dx$$

Answered by mathmax by abdo last updated on 12/Feb/21

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}\mathrm{I}={\int }_0^1\frac{\ln (\mathrm{x}+1)}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}$$$$\mathrm{also}{\ln }^{{}^{\prime }}(1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \ln (1+\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \frac{\ln (1+\mathrm{x})}{\mathrm{x}}.\frac{1}{1+{\mathrm{x}}^2}=\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\right)\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\right)$$$$=\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}{\mathrm{x}}^{\mathrm{n}}\right).\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\right)=\left(\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{n}}\right).\left(\mathrm{\Sigma }{\mathrm{b}}_{\mathrm{n}}\right)$$$$=\mathrm{\Sigma }{\mathrm{c}}_{\mathrm{n}}\mathrm{with}{\mathrm{c}}_{\mathrm{n}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{b}}_{\mathrm{j}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}+1}{\mathrm{x}}^{\mathrm{i}}{(-1)}^{\mathrm{j}}{\mathrm{x}}^{2\mathrm{j}}$$$$=\sum _{\mathrm{i}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}+1}{\mathrm{x}}^{\mathrm{i}}{(-1)}^{\mathrm{n}-\mathrm{i}}{\mathrm{x}}^{2(\mathrm{n}-\mathrm{i})}\Rightarrow$$$$\mathrm{I}={\int }_0^1\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}\mathrm{dx}={\int }_0^1\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{i}+1}{\mathrm{x}}^{2\mathrm{n}-\mathrm{i}}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\left(\sum _{\mathrm{i}=0}^{\mathrm{n}}\frac{1}{\mathrm{i}+1}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}-\mathrm{i}}\mathrm{dx}\right)$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\left(\sum _{\mathrm{i}=0}^{\mathrm{n}}\frac{1}{(\mathrm{i}+1)(2\mathrm{n}-\mathrm{i}+1)}\right)$$

Answered by mathmax by abdo last updated on 12/Feb/21

$$\mathrm{let}\mathrm{try}\mathrm{parametric}\mathrm{method}\mathrm{\Phi }={\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}\mathrm{let}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^1\frac{\ln (1+\mathrm{ax})}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{with}\mathrm{a}>0\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^1\frac{\mathrm{x}}{(1+\mathrm{ax})\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}$$$$={\int }_0^1\frac{\mathrm{dx}}{(1+\mathrm{ax})({\mathrm{x}}^2+1)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{ax}+1)({\mathrm{x}}^2+1)}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\alpha }{\mathrm{ax}+1}+\frac{\mathrm{mx}+\mathrm{n}}{{\mathrm{x}}^2+1}$$$$\alpha =\frac{1}{(\frac{1}{{\mathrm{a}}^2}+1)}=\frac{{\mathrm{a}}^2}{1+{\mathrm{a}}^2},{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\frac{\alpha }{\mathrm{a}}+\mathrm{m}\Rightarrow \mathrm{m}=-\frac{\alpha }{\mathrm{a}}=-\frac{\mathrm{a}}{1+{\mathrm{a}}^2}$$$$\mathrm{F}\left(0\right)=1=\alpha +\mathrm{n}\Rightarrow \mathrm{n}=1-\alpha =1-\frac{{\mathrm{a}}^2}{1+{\mathrm{a}}^2}=\frac{1}{1+{\mathrm{a}}^2}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{{\mathrm{a}}^2}{(1+{\mathrm{a}}^2)(\mathrm{ax}+1)}+\frac{-\frac{\mathrm{a}}{1+{\mathrm{a}}^2}\mathrm{x}+\frac{1}{1+{\mathrm{a}}^2}}{{\mathrm{x}}^2+1}$$$$\Rightarrow {\int }_0^1\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{{\mathrm{a}}^2}{1+{\mathrm{a}}^2}{\int }_0^1\frac{\mathrm{dx}}{\mathrm{ax}+1}-\frac{1}{1+{\mathrm{a}}^2}{\int }_0^1\frac{\mathrm{ax}-1}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=\frac{\mathrm{a}}{1+{\mathrm{a}}^2}{\left[\ln (\mathrm{ax}+1)\right]}_0^1-\frac{\mathrm{a}}{2(1+{\mathrm{a}}^2)}{\left[\ln (1+{\mathrm{x}}^2)\right]}_0^1+\frac{1}{1+{\mathrm{a}}^2}.\frac{\pi }{4}$$$$=\frac{\mathrm{a}}{1+{\mathrm{a}}^2}\ln (\mathrm{a}+1)-\frac{\mathrm{aln}\left(2\right)}{2(1+{\mathrm{a}}^2)}+\frac{\pi }{4(1+{\mathrm{a}}^2)}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\int \frac{\mathrm{aln}(\mathrm{a}+1)}{1+{\mathrm{a}}^2}\mathrm{da}-\frac{\ln \left(2\right)}{2}\int \frac{\mathrm{ada}}{1+{\mathrm{a}}^2}+\frac{\pi }{4}\mathrm{arctana}+\mathrm{c}$$$$=\int \frac{\mathrm{aln}(1+\mathrm{a})}{1+{\mathrm{a}}^2}\mathrm{da}-\frac{\ln \left(2\right)}{4}\ln (1+{\mathrm{a}}^2)+\frac{\pi }{4}\arctan \left(\mathrm{a}\right)+\mathrm{C}$$$$\int \frac{\mathrm{a}}{{\mathrm{a}}^2+1}\ln (1+\mathrm{a})\mathrm{da}=\frac{1}{2}\ln ({\mathrm{a}}^2+1)\ln (1+\mathrm{a})-\int \frac{1}{2}\ln ({\mathrm{a}}^2+1)\frac{\mathrm{da}}{1+\mathrm{a}}$$$$=\frac{1}{2}\ln (1+\mathrm{a})\ln (1+{\mathrm{a}}^2)-\frac{1}{2}\int \frac{\ln (1+{\mathrm{a}}^2)}{1+\mathrm{a}}\mathrm{da}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{a}}\frac{\mathrm{xln}(1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}-\frac{\ln \left(2\right)}{4}\ln (1+{\mathrm{a}}^2)+\frac{\pi }{4}\arctan \left(\mathrm{a}\right)+\mathrm{C}$$$$\mathrm{C}=\mathrm{f}\left(0\right)=0$$$$\mathrm{f}\left(1\right)={\int }_0^1\frac{\mathrm{xln}(1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}-\frac{{\ln }^2\left(2\right)}{4}+\frac{{\pi }^2}{16}=\mathrm{\Phi }$$$$\mathrm{we}\mathrm{have}\mathrm{by}\mathrm{parts}{\int }_0^1\frac{\mathrm{xln}(1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}={\left[\frac{1}{2}\ln (1+\mathrm{x})\ln (1+{\mathrm{x}}^2)\right]}_0^1$$$$-\frac{1}{2}{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+\mathrm{x}}\mathrm{dx}=\frac{1}{2}{\ln }^2\left(2\right)-\frac{1}{2}{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+\mathrm{x}}\mathrm{dx}$$$${\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+\mathrm{x}}\mathrm{dx}={\int }_0^1\ln (1+{\mathrm{x}}^2)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln (1+{\mathrm{x}}^2)\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln (1+{\mathrm{x}}^2)\mathrm{dx}={\left[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\ln (1+{\mathrm{x}}^2)\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\times \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{\ln \left(2\right)}{\mathrm{n}+1}-\frac{2}{\mathrm{n}+1}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+2}}{1+{\mathrm{x}}^2}\mathrm{dx}.....\mathrm{be}\mathrm{continued}...$$

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