2-e-1-ln-x-1-ln-2-x-dx-

Question Number 87669 by john santu last updated on 05/Apr/20

$$\int_{\mathrm{2}} ^{\:\:\mathrm{e}} \left(\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \mathrm{x}}\right)\:\mathrm{dx}? \\ $$

Commented by mathmax by abdo last updated on 05/Apr/20

I =∫_2 ^e (((lnx−1)/(ln^2 x)))dx vhangement lnx =t give I =∫_(ln2) ^1 ((t−1)/t^2 )e^t dt =∫_(ln2) ^1 (((t−1)e^t )/t^2 )dt =_(by parts) [−(1/t)(t−1)e^t ]_(ln(2)) ^1 +∫_(ln(2)) ^1 (1/t){ e^t +(t−1)e^t }dt =(((ln2−1)2)/(ln2)) +∫_(ln2) ^1 e^t dt =2−(2/(ln2)) +e−2 =e−(2/(ln2))

$${I}\:=\int_{\mathrm{2}} ^{{e}} \left(\frac{{lnx}−\mathrm{1}}{{ln}^{\mathrm{2}} {x}}\right){dx}\:{vhangement}\:{lnx}\:={t}\:{give} \\ $$$${I}\:=\int_{{ln}\mathrm{2}} ^{\mathrm{1}} \:\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} }{e}^{{t}} \:{dt}\:\:=\int_{{ln}\mathrm{2}} ^{\mathrm{1}} \:\frac{\left({t}−\mathrm{1}\right){e}^{{t}} }{{t}^{\mathrm{2}} }{dt}\:=_{{by}\:{parts}} \:\left[−\frac{\mathrm{1}}{{t}}\left({t}−\mathrm{1}\right){e}^{{t}} \right]_{{ln}\left(\mathrm{2}\right)} ^{\mathrm{1}} \\ $$$$+\int_{{ln}\left(\mathrm{2}\right)} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{t}}\left\{\:\:{e}^{{t}} \:+\left({t}−\mathrm{1}\right){e}^{{t}} \right\}{dt}\:=\frac{\left({ln}\mathrm{2}−\mathrm{1}\right)\mathrm{2}}{{ln}\mathrm{2}}\:+\int_{{ln}\mathrm{2}} ^{\mathrm{1}} {e}^{{t}} \:{dt} \\ $$$$=\mathrm{2}−\frac{\mathrm{2}}{{ln}\mathrm{2}}\:+{e}−\mathrm{2}\:={e}−\frac{\mathrm{2}}{{ln}\mathrm{2}} \\ $$

Commented by Ar Brandon last updated on 05/Apr/20

It seems you made an error while integrating by part. It′s not −(1/t)(t−1)e^t but (lnt+(1/t))e^t May be you should verify your solution.

$${It}\:{seems}\:{you}\:{made}\:{an}\:{error}\:{while}\:{integrating}\:{by}\:{part}. \\ $$$${It}'{s}\:{not}\:−\frac{\mathrm{1}}{{t}}\left({t}−\mathrm{1}\right){e}^{{t}} \:{but}\:\left({lnt}+\frac{\mathrm{1}}{{t}}\right){e}^{{t}} \\ $$$${May}\:{be}\:{you}\:{should}\:{verify}\:{your}\:{solution}. \\ $$

Commented by abdomathmax last updated on 05/Apr/20

no sir by psrts i have taken u^′ =(1/t^2 ) and v=(t−1)e^t

$${no}\:{sir}\:{by}\:{psrts}\:{i}\:{have}\:{taken}\:{u}^{'} \:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${and}\:{v}=\left({t}−\mathrm{1}\right){e}^{{t}} \:\:\: \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 05/Apr/20

$${OK}\:{bro}.\:{It}'{s}\:{clear},\:{thanks}. \\ $$

Answered by TANMAY PANACEA. last updated on 05/Apr/20

∫_2 ^e ((lnx−1)/((lnx)^2 ))dx ∫_2 ^e ((lnx×(dx/dx)−x×(d/dx)(lnx))/((lnx)^2 ))dx ∫_2 ^e (d/dx)((x/(lnx)))dx ∫_2 ^e d((x/(lnx)))=∣(x/(lnx))∣_2 ^e =(e/(lne))−(2/(ln2))=e−(2/(ln2))(corrected)

$$\int_{\mathrm{2}} ^{{e}} \frac{{lnx}−\mathrm{1}}{\left({lnx}\right)^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{2}} ^{{e}} \frac{{lnx}×\frac{{dx}}{{dx}}−{x}×\frac{{d}}{{dx}}\left({lnx}\right)}{\left({lnx}\right)^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{2}} ^{{e}} \frac{{d}}{{dx}}\left(\frac{{x}}{{lnx}}\right){dx} \\ $$$$\int_{\mathrm{2}} ^{{e}} {d}\left(\frac{{x}}{{lnx}}\right)=\mid\frac{{x}}{{lnx}}\mid_{\mathrm{2}} ^{{e}} =\frac{{e}}{{lne}}−\frac{\mathrm{2}}{{ln}\mathrm{2}}={e}−\frac{\mathrm{2}}{{ln}\mathrm{2}}\left({corrected}\right) \\ $$

Commented by TANMAY PANACEA. last updated on 05/Apr/20