Question Number 151780 by Tawa11 last updated on 23/Aug/21
Answered by puissant last updated on 23/Aug/21
$${Q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctanx}}{{x}}{dx} \\ $$$${arctanx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\frac{{arctanx}}{{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctanx}}{{x}}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} {dx} \\ $$$$=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by Tawa11 last updated on 23/Aug/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$