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Question Number 106098 by redmiiuser last updated on 02/Aug/20

$${\int }_{\frac{1}{2014}}^{2014}\frac{{\tan }^{-1}x}{x}dx$$

Answered by abdomathmax last updated on 02/Aug/20

$$\mathrm{let}\mathrm{find}\mathrm{f}\left(\mathrm{a}\right)={\int }_{\frac{1}{\mathrm{a}}}^{\mathrm{a}}\frac{\mathrm{arctanx}}{\mathrm{x}}\mathrm{dx}\mathrm{with}\mathrm{a}>0$$$$\mathrm{changement}\mathrm{x}=\frac{1}{\mathrm{t}}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{a}\right)=-{\int }_{\frac{1}{\mathrm{a}}}^{\mathrm{a}}\frac{\arctan \left(\frac{1}{\mathrm{t}}\right)}{{\mathrm{t}}^2}(-\mathrm{t})\mathrm{dt}$$$$={\int }_{\frac{1}{\mathrm{a}}}^{\mathrm{a}}\frac{\frac{\pi }{2}-\arctan \left(\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}=\frac{\pi }{2}{\int }_{\frac{1}{\mathrm{a}}}^{\mathrm{a}}\frac{\mathrm{dt}}{\mathrm{t}}-\mathrm{f}\left(\mathrm{a}\right)\Rightarrow$$$$2\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}{[\ln \mid \mathrm{t}\mid ]}_{\frac{1}{\mathrm{a}}}^{\mathrm{a}}=\frac{\pi }{2}\{\ln \left(\mathrm{a}\right)+\ln \left(\mathrm{a}\right)\}=\pi \ln \left(\mathrm{a}\right)$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}\ln \left(\mathrm{a}\right)\Rightarrow {\int }_{\frac{1}{2014}}^{2014}\frac{\mathrm{arctanx}}{\mathrm{x}}\mathrm{dx}=\frac{\pi }{2}\ln \left(2014\right)$$

Commented by redmiiuser last updated on 02/Aug/20

$$perfect$$

Commented by Aziztisffola last updated on 02/Aug/20

$$\mathrm{Nice}\mathrm{method}\mathrm{Sir}.$$

Commented by mathmax by abdo last updated on 02/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}.$$

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