Question Number 64559 by aliesam last updated on 19/Jul/19
Commented by mathmax by abdo last updated on 20/Jul/19
![let I =∫_(1/e) ^e (√((1−(lnx)^2 )/x))dx ⇒I =∫_(1/e) ^e (√(1−(lnx)^2 ))(dx/( (√x))) changement (√x)=t give I = ∫_(1/( (√e))) ^(√e) (√(1−(2lnt)^2 )) ((2tdt)/t) =2 ∫_(1/( (√e))) ^(√e) (√(1−4ln^2 t)) dt changement t =(1/u) give I =2 ∫_(√e) ^(1/( (√e))) (√(1−4ln^2 u))(−(du/u^2 )) =2 ∫_(1/( (√e))) ^(√e) (1/u^2 )(√(1−4ln^2 u))du by psrts f′=(1/u^2 ) and g =(√(1−4ln^2 u)) ⇒ I =2{ [−(1/u)(√(1−4ln^2 u))]_(√e) ^(1/( (√e))) −∫_(1/( (√e))) ^(√e) (−(1/u)) ((−8lnu)/(2u(√(1−4ln^2 u)))) du} =2{....−4 ∫_(1/( (√e))) ^(√e) ((lnu)/(u^2 (√(1−4ln^2 u)))) du}....be continued...](https://www.tinkutara.com/question/Q64643.png)
$${let}\:{I}\:=\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} \sqrt{\frac{\mathrm{1}−\left({lnx}\right)^{\mathrm{2}} }{{x}}}{dx}\:\Rightarrow{I}\:=\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} \sqrt{\mathrm{1}−\left({lnx}\right)^{\mathrm{2}} }\frac{{dx}}{\:\sqrt{{x}}} \\ $$$${changement}\:\sqrt{{x}}={t}\:{give}\:\:{I}\:=\:\int_{\frac{\mathrm{1}}{\:\sqrt{{e}}}} ^{\sqrt{{e}}} \:\:\sqrt{\mathrm{1}−\left(\mathrm{2}{lnt}\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{tdt}}{{t}} \\ $$$$=\mathrm{2}\:\int_{\frac{\mathrm{1}}{\:\sqrt{{e}}}} ^{\sqrt{{e}}} \:\:\sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {t}}\:{dt}\:\:\:\:{changement}\:{t}\:=\frac{\mathrm{1}}{{u}}\:{give} \\ $$$${I}\:=\mathrm{2}\:\int_{\sqrt{{e}}} ^{\frac{\mathrm{1}}{\:\sqrt{{e}}}} \sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {u}}\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right)\:=\mathrm{2}\:\int_{\frac{\mathrm{1}}{\:\sqrt{{e}}}} ^{\sqrt{{e}}} \frac{\mathrm{1}}{{u}^{\mathrm{2}} }\sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {u}}{du} \\ $$$${by}\:{psrts}\:\:{f}'=\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\:{and}\:{g}\:=\sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {u}}\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\left\{\:\:\left[−\frac{\mathrm{1}}{{u}}\sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {u}}\right]_{\sqrt{{e}}} ^{\frac{\mathrm{1}}{\:\sqrt{{e}}}} \:\:−\int_{\frac{\mathrm{1}}{\:\sqrt{{e}}}} ^{\sqrt{{e}}} \left(−\frac{\mathrm{1}}{{u}}\right)\:\:\frac{−\mathrm{8}{lnu}}{\mathrm{2}{u}\sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {u}}}\:{du}\right\} \\ $$$$=\mathrm{2}\left\{….−\mathrm{4}\:\int_{\frac{\mathrm{1}}{\:\sqrt{{e}}}} ^{\sqrt{{e}}} \:\:\frac{{lnu}}{{u}^{\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{4}{ln}^{\mathrm{2}} {u}}}\:{du}\right\}….{be}\:{continued}… \\ $$