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
![∫_5 ^(1/5) ((arctanx)/x)dx =[lnx∙arctanx]_5 ^(1/5) −∫_5 ^(1/5) ((lnx)/(1+x^2 ))dx =−ln5(arctan((1/5))+arctan(5))−∫_5 ^(1/5) ((lnx)/(1+x^2 ))dx I=∫_5 ^(1/5) ((lnx)/(1+x^2 ))dx , x=(1/t) ⇒dx=−(1/t^2 )dt I=∫_(1/5) ^5 ((lnt)/(1+(1/t^2 )))∙(1/t^2 )dt=∫_(1/5) ^5 ((lnt)/(1+t^2 ))dt=−∫_5 ^(1/5) ((lnt)/(1+t^2 ))dt 2I=∫_5 ^(1/5) ((lnx)/(1+x^2 ))dx−∫_5 ^(1/5) ((lnt)/(1+t^2 ))dt=0 ⇒I=0 ⇒∫_5 ^(1/5) ((arctanx)/x)dx=−ln5(arctan((1/5))+arctan(5))=−(π/2)ln5](https://www.tinkutara.com/question/Q172591.png)
$$\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{arctan}{x}}{{x}}{dx} \\ $$$$\:\:\:=\left[\mathrm{ln}{x}\centerdot\mathrm{arctan}{x}\right]_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} −\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=−\mathrm{ln5}\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{arctan}\left(\mathrm{5}\right)\right)−\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}=\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:,\:{x}=\frac{\mathrm{1}}{{t}}\:\Rightarrow{dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$${I}=\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \frac{\mathrm{ln}{t}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\centerdot\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt}=\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=−\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{0}\:\Rightarrow{I}=\mathrm{0} \\ $$$$\Rightarrow\int_{\mathrm{5}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{\mathrm{arctan}{x}}{{x}}{dx}=−\mathrm{ln5}\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{arctan}\left(\mathrm{5}\right)\right)=−\frac{\pi}{\mathrm{2}}\mathrm{ln5} \\ $$
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
![∫_(1/5) ^5 ((arctanx)/x)dx=k posons { ((u=arctanx)),((v′=(1/x))) :}=> { ((u′=(1/(1+x^2 )))),((v=ln(x))) :} =[arctanx×lmx]_(1/5) ^5 −∫_(1/5) ^5 ((lnx)/(1+x^2 ))dx =ln(5)(arctan(5)+arctan((1/5)))−∫_(1/5) ^5 ((lnx)/(1+x^2 ))dx posonsM=∫_(1/5) ^5 ((lnx)/(1+x^2 ))dx=? posons x=(1/t)=>dx=−(1/t^2 )dt qd: { ((x−>5)),((x−>(1/5))) :}=> { ((t−>(1/5))),((t→5)) :} =∫_5 ^(1/5) −((ln((1/t)))/(1+(1/t^2 )))×(1/t^2 )dt =−∫_(1/5) ^5 ((lnt)/(1+t^2 ))dt =>∫_(1/5) ^5 ((lnx)/(1+x^2 ))dx=0 =>∫_(1/5) ^5 ((arctanx)/x)dx=ln5(arctan5+arctan((1/5))) or ∀x∈R^∗ on a arctan(x)+arctan((1/x))=(𝛑/2) =>∫_(1/5_ ) ^5 ((arctanx)/(1+x^2 ))dx=(𝛑/2)ln5 .........Le ce^ le^ bre cedric junior..............](https://www.tinkutara.com/question/Q172748.png)
$$\int_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} \frac{\boldsymbol{{arctanx}}}{\boldsymbol{{x}}}\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{posons}}\:\begin{cases}{\boldsymbol{{u}}=\boldsymbol{{arctanx}}}\\{\boldsymbol{{v}}'=\frac{\mathrm{1}}{\boldsymbol{{x}}}}\end{cases}=>\begin{cases}{\boldsymbol{{u}}'=\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\\{\boldsymbol{{v}}=\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)}\end{cases} \\ $$$$=\left[\boldsymbol{{arctanx}}×\boldsymbol{{lmx}}\right]_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} −\int_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} \frac{\boldsymbol{\mathrm{lnx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}} \\ $$$$=\boldsymbol{\mathrm{ln}}\left(\mathrm{5}\right)\left(\boldsymbol{\mathrm{arctan}}\left(\mathrm{5}\right)+\boldsymbol{\mathrm{arctan}}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\right)−\int_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} \frac{\boldsymbol{\mathrm{lnx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posonsM}}=\int_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} \frac{\boldsymbol{\mathrm{lnx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=? \\ $$$$\boldsymbol{\mathrm{posons}}\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}}=>\boldsymbol{\mathrm{dx}}=−\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\boldsymbol{\mathrm{dt}} \\ $$$$\boldsymbol{\mathrm{qd}}:\begin{cases}{\boldsymbol{\mathrm{x}}−>\mathrm{5}}\\{\boldsymbol{\mathrm{x}}−>\frac{\mathrm{1}}{\mathrm{5}}}\end{cases}=>\begin{cases}{\boldsymbol{\mathrm{t}}−>\frac{\mathrm{1}}{\mathrm{5}}}\\{\boldsymbol{\mathrm{t}}\rightarrow\mathrm{5}}\end{cases} \\ $$$$=\int_{\mathrm{5}} ^{\mathrm{1}/\mathrm{5}} −\frac{\boldsymbol{\mathrm{ln}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}^{\mathrm{2}} }}×\frac{\mathrm{1}}{\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\boldsymbol{\mathrm{dt}} \\ $$$$=−\int_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} \frac{\boldsymbol{\mathrm{lnt}}}{\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\boldsymbol{\mathrm{dt}} \\ $$$$=>\int_{\mathrm{1}/\mathrm{5}} ^{\mathrm{5}} \frac{\boldsymbol{\mathrm{lnx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\mathrm{0} \\ $$$$=>\int_{\frac{\mathrm{1}}{\mathrm{5}}} ^{\mathrm{5}} \frac{\boldsymbol{{arctanx}}}{\boldsymbol{{x}}}\boldsymbol{{dx}}=\boldsymbol{{ln}}\mathrm{5}\left(\boldsymbol{{arctan}}\mathrm{5}+\boldsymbol{{arctan}}\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\right) \\ $$$$\boldsymbol{{or}}\:\forall\boldsymbol{{x}}\in\mathbb{R}^{\ast} \:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}} \\ $$$$\boldsymbol{\mathrm{arctan}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{arctan}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$=>\int_{\frac{\mathrm{1}}{\mathrm{5}_{} }} ^{\mathrm{5}} \frac{\boldsymbol{\mathrm{arctanx}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\mathrm{5} \\ $$$$………\mathscr{L}\boldsymbol{\mathrm{e}}\:\boldsymbol{\mathrm{c}}\acute {\boldsymbol{\mathrm{e}l}}\grave {\boldsymbol{\mathrm{e}bre}}\:\boldsymbol{\mathrm{cedric}}\:\boldsymbol{\mathrm{junior}}………….. \\ $$$$ \\ $$