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Question Number 91220 by Ar Brandon last updated on 28/Apr/20

$${\int }_1^{\mathrm{x}}\frac{\mathrm{lnt}}{1+{\mathrm{t}}^2}\mathrm{dt}$$

Commented by abdomathmax last updated on 28/Apr/20

$$lettakeatry$$$$ifx>1wedothechangementt=\frac{1}{u}\Rightarrow$$$$I=-{\int }_{\frac{1}{x}}^1\frac{-lnu}{1+\frac{1}{u^2}}(-\frac{du}{u^2})=-{\int }_{\frac{1}{x}}^1\frac{lnu}{1+u^2}du$$$$=-{\int }_{\frac{1}{x}}^{1}lnu\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{u}^{2n}\right)du$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^{n+1}{\int }_{\frac{1}{x}}^1{u}^{2n}lnudu=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^{n+1}{U}_n$$$$U_n={\int }_{\frac{1}{x}}^1{u}^{2n}ln\left(u\right)du{=}_{byparts}{\left[\frac{u^{2n+1}}{2n+1}lnu\right]}_{\frac{1}{x}}^1$$$$-{\int }_{\frac{1}{x}}^1\frac{u^{2n}}{2n+1}du=\frac{1}{2n+1}\frac{lnx}{x^{2n+1}}-\frac{1}{2n+1}{\left[\frac{1}{2n+1}{u}^{2n+1}\right]}_{\frac{1}{x}}^1$$$$=\frac{ln\left(x\right)}{(2n+1)x^{2n+1}}-\frac{1}{{(2n+1)}^2}(1-\frac{1}{x^{2n+1}})\Rightarrow$$$$I=-lnx\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)x^{2n+1}}+\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}$$$$-\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2{x}^{2n+1}}resttocalculatethosesums$$$$ifx<1I=-{\int }_x^1\frac{lnt}{1+t^2}dt$$$$=-{\int }_x^{1}lnt\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{2n}\right)dt$$$$=-\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_x^1{t}^{2n}lntdtandwefollowthe$$$$sameway....becontinued...$$

Commented by Ar Brandon last updated on 28/Apr/20

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Commented by mathmax by abdo last updated on 29/Apr/20

$$thankx$$

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