Question Number 172555 by cortano1 last updated on 28/Jun/22 | ||
$$\:\:\:\:\:\:\underset{\mathrm{1}/\mathrm{5}} {\overset{\mathrm{5}} {\int}}\:\frac{\mathrm{arctan}\:\left({x}\right)}{{x}}\:{dx}\:=? \\ $$ | ||
Commented by greougoury555 last updated on 28/Jun/22 | ||
$$\:=\:\mathrm{ln}\:\sqrt{\mathrm{5}^{\pi} }\:=\:\frac{\pi}{\mathrm{2}}\:\mathrm{ln}\:\mathrm{5} \\ $$ | ||
Commented by cortano1 last updated on 29/Jun/22 | ||
Commented by shikaridwan last updated on 29/Jun/22 | ||
$${Generally}\:\int_{\mathrm{1}/{a}} ^{{a}} \frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\left({a}\right) \\ $$ | ||
Answered by Mathspace last updated on 28/Jun/22 | ||
$$\Upsilon=\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \frac{{arctanx}}{{x}}{dx} \\ $$$$\Upsilon=_{{x}=\frac{\mathrm{1}}{{t}}} \:\:\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{−{arctan}\left({t}\right)}{\frac{\mathrm{1}}{{t}}}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right) \\ $$$$=−\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \:\:\frac{{arctant}}{{t}}{dt}=−\Upsilon\:\Rightarrow \\ $$$$\mathrm{2}\Upsilon=\mathrm{0}\:\Rightarrow\Upsilon=\mathrm{0} \\ $$ | ||
Commented by cortano1 last updated on 29/Jun/22 | ||
$${wrong} \\ $$ | ||
Commented by cortano1 last updated on 29/Jun/22 | ||
Answered by Ar Brandon last updated on 29/Jun/22 | ||
$$\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{arctan}{x}}{{x}}{dx} \\ $$$$\:\:\:=\left[\mathrm{ln}{x}\centerdot\mathrm{arctan}{x}\right]_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} −\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=−\mathrm{ln5}\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{arctan}\left(\mathrm{5}\right)\right)−\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}=\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:,\:{x}=\frac{\mathrm{1}}{{t}}\:\Rightarrow{dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$${I}=\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \frac{\mathrm{ln}{t}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\centerdot\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt}=\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=−\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{0}\:\Rightarrow{I}=\mathrm{0} \\ $$$$\Rightarrow\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{arctan}{x}}{{x}}{dx}=−\mathrm{ln5}\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{arctan}\left(\mathrm{5}\right)\right)=−\frac{\pi}{\mathrm{2}}\mathrm{ln5} \\ $$ | ||
Commented by shikaridwan last updated on 29/Jun/22 | ||
$$ \\ $$🥲. | ||
Commented by shikaridwan last updated on 29/Jun/22 | ||
$${hey}! \\ $$ | ||
Commented by Ar Brandon last updated on 29/Jun/22 | ||
Hey bro Are you back for real? I missed you so much | ||
Commented by Ar Brandon last updated on 29/Jun/22 | ||
I tried reaching you inbox but couldn't. | ||
Commented by Rasheed.Sindhi last updated on 29/Jun/22 | ||
Happy to see you both meeting in the forum once again! 🌷🌺🌹🥀 | ||
Commented by Ar Brandon last updated on 29/Jun/22 | ||
Thanks Sir | ||
Answered by CElcedricjunior last updated on 01/Jul/22 | ||