Question Number 144143 by ArielVyny last updated on 22/Jun/21
$$\int_{\frac{\mathrm{1}}{{a}}} ^{{a}} \frac{{arctg}\left({x}\right)}{{x}}{dx}=??? \\ $$
Answered by mathmax by abdo last updated on 22/Jun/21
![I=∫_(1/a) ^a ((arctanx)/x)dx ⇒I=_(x=(1/t)) −∫_(1/a) ^a (((π/2)−arctant)/(1/t))(−(dt/t^2 )) =(π/2)∫_(1/a) ^a (dt/t) −∫_(1/a) ^a ((arctant)/t)dt =(π/2)[log∣t∣]_(1/a) ^a −I ⇒ 2I=(π/2)(2log∣a∣) ⇒I=(π/2)log∣a∣](https://www.tinkutara.com/question/Q144167.png)
$$\mathrm{I}=\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\mathrm{arctanx}}{\mathrm{x}}\mathrm{dx}\:\Rightarrow\mathrm{I}=_{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}} \:\:\:−\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{arctant}}{\frac{\mathrm{1}}{\mathrm{t}}}\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \frac{\mathrm{dt}}{\mathrm{t}}\:−\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\mathrm{arctant}}{\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{2}}\left[\mathrm{log}\mid\mathrm{t}\mid\right]_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:−\mathrm{I}\:\Rightarrow \\ $$$$\mathrm{2I}=\frac{\pi}{\mathrm{2}}\left(\mathrm{2log}\mid\mathrm{a}\mid\right)\:\Rightarrow\mathrm{I}=\frac{\pi}{\mathrm{2}}\mathrm{log}\mid\mathrm{a}\mid \\ $$