Question Number 84570 by msup trace by abdo last updated on 14/Mar/20
$${calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{arctan}\left(\frac{\mathrm{3}}{{x}}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 15/Mar/20
![let I =∫_1 ^(+∞) ((arctan((3/x)))/x^2 )dx changement (3/x) =t give (x/3) =(1/t) ⇒ x =(3/t) ⇒ I =∫_3 ^0 ((arctan(t))/(((3/t))^2 ))(−(3/t^2 ))dt =(1/3)∫_0 ^3 arctant dt =_(by parts) =(1/3){ [tarctant]_0 ^3 −∫_0 ^3 t(dt/(1+t^2 ))} 3I =3arctan(3) −(1/2)[ln(1+t^2 )]_0 ^3 =3arctan(3)−(1/2)ln(10) ⇒I =arctan(3)−(1/6)ln(10)](https://www.tinkutara.com/question/Q84707.png)
$${let}\:{I}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{arctan}\left(\frac{\mathrm{3}}{{x}}\right)}{{x}^{\mathrm{2}} }{dx}\:{changement}\:\frac{\mathrm{3}}{{x}}\:={t}\:{give}\:\frac{{x}}{\mathrm{3}}\:=\frac{\mathrm{1}}{{t}}\:\Rightarrow \\ $$$${x}\:=\frac{\mathrm{3}}{{t}}\:\Rightarrow\:{I}\:=\int_{\mathrm{3}} ^{\mathrm{0}} \:\frac{{arctan}\left({t}\right)}{\left(\frac{\mathrm{3}}{{t}}\right)^{\mathrm{2}} }\left(−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{3}} \:\:{arctant}\:{dt}\:=_{{by}\:{parts}} \:=\frac{\mathrm{1}}{\mathrm{3}}\left\{\:\:\left[{tarctant}\right]_{\mathrm{0}} ^{\mathrm{3}} \:−\int_{\mathrm{0}} ^{\mathrm{3}} \:{t}\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right\} \\ $$$$\mathrm{3}{I}\:=\mathrm{3}{arctan}\left(\mathrm{3}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$=\mathrm{3}{arctan}\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{10}\right)\:\Rightarrow{I}\:={arctan}\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\mathrm{10}\right) \\ $$
Answered by M±th+et£s last updated on 14/Mar/20
![y=(3/x) dy=((−3)/x^2 ) I=∫_3 ^0 tan^(−1) (y) (dy/(−3)) =(1/3)∫_0 ^3 tan^(−1) y dy =(1/3)[y tan^(−1) y − (1/2)ln(y^2 +1)]_(0 ) ^3 (1/3)(tan^(−1) (x)−(1/2)ln(10))](https://www.tinkutara.com/question/Q84615.png)
$${y}=\frac{\mathrm{3}}{{x}}\:{dy}=\frac{−\mathrm{3}}{{x}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{3}} ^{\mathrm{0}} {tan}^{−\mathrm{1}} \left({y}\right)\:\frac{{dy}}{−\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{3}} {tan}^{−\mathrm{1}} {y}\:{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[{y}\:{tan}^{−\mathrm{1}} {y}\:−\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}\:} ^{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left({tan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{10}\right)\right) \\ $$