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Question Number 128721 by rs4089 last updated on 09/Jan/21

$${\int }_0^1\frac{\ln \mathrm{x}}{x(x^2+1)}\mathrm{dx}$$

Commented by Dwaipayan Shikari last updated on 09/Jan/21

$$\frac{1}{2}{\int }_0^1\frac{logx}{x}(\frac{1}{x-i}+\frac{1}{x+i})dx$$$$=\frac{1}{2i}{\int }_0^1\frac{logx}{x^2}(\frac{xi}{x-i}+\frac{xi}{x+i})dx$$$$=\frac{1}{2i}{\int }_0^1\frac{logx}{x^2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left({xe}^{\frac{\pi }{2}i}\right)}^n-{\left({xe}^{-\frac{\pi }{2}i}\right)}^n$$$$={\int }_0^1\frac{logx}{x^2}.\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^n\frac{sin\left(\frac{\pi }{2}n\right)}{n^2}=\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{sin\left(\frac{\pi }{2}n\right)}{n^2}{\int }_0^1{x}^{n-2}log\left(x\right)dx$$$$=(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+...){\int }_0^{\mathrm{\infty }}{e}^{-(n-1)t}tdt=G\left(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n-1)}^2}\right)\to \mathrm{\infty }$$

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