Question Number 189975 by mathlove last updated on 25/Mar/23
Answered by a.lgnaoui last updated on 27/Mar/23
![Q_1 ((lnx)/(2−x))=−((lnx)/(x(1−(2/x))))=((lnx)/x)×((x/(2−x))) U^′ =((lnx)/x) →U=(1/2)(lnx)^2 V=(x/(2−x)) →V′=(2/((2−x)^2 )) UV=((x(lnx)^2 )/(2(2−x))) UV^′ =(((lnx)^2 )/((x−2)^2 )) =(x^2 /((x−2)^2 ))×(((lnx)/x))^2 =(((lnx)^2 )/((x−2)^2 )) I=UV−∫UV^′ =((x(lnx)^2 )/(2(x−2)))−∫(([ln((x−2)+2)]^2 )/((x−2)^2 )) x−2=t dx=dt −∫(([ln(t+2)]^2 )/t^2 )dt u^′ =−(1/t^2 ) → u=(1/t) v=[ln(t+2)]^2 →v′=2ln(t+2)(1/(t+2)) uv=((ln(t+2))/t) uv′=((2ln(t+2))/t) =2((lnx)/(x−2)) I=((x(lnx)^2 )/(2(x−2)))+((lnx)/(x−2))−2∫((lnx)/(x−2)) 3I=[((x(lnx)^2 )/(2(x−2)))]_(3/2) ^1 +[((lnx)/(x−2))]_(3/2) ^1 I=(1/3)[(−(((3/2)(ln3−ln2)^2 )/(2((3/2)−2))))−((ln3−ln2)/((3/2)−2))] =(1/3)(((3()/)((ln3−ln2)^2 )/2))+2×((ln3−ln2)/3) =(((ln3−ln2)^2 )/2)+((2(ln3−ln2))/3) I=((3(ln3)^2 +3(ln2)^2 −4(ln3−ln2))/6)](https://www.tinkutara.com/question/Q190085.png)
$${Q}_{\mathrm{1}} \:\:\:\frac{{lnx}}{\mathrm{2}−{x}}=−\frac{{lnx}}{{x}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)}=\frac{{lnx}}{{x}}×\left(\frac{{x}}{\mathrm{2}−{x}}\right) \\ $$$${U}^{'} =\frac{{lnx}}{{x}}\:\:\:\:\:\:\:\:\:\:\rightarrow{U}=\frac{\mathrm{1}}{\mathrm{2}}\left({lnx}\right)^{\mathrm{2}} \\ $$$${V}=\frac{{x}}{\mathrm{2}−{x}}\:\:\:\:\:\:\:\:\:\:\rightarrow{V}'=\frac{\mathrm{2}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\ $$$${UV}=\frac{{x}\left({lnx}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}−{x}\right)}\:\:\:\:\:\:\:{UV}^{'} =\frac{\left({lnx}\right)^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }×\left(\frac{{lnx}}{{x}}\right)^{\mathrm{2}} =\frac{\left({lnx}\right)^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${I}={UV}−\int{UV}^{'} \\ $$$$=\frac{{x}\left({lnx}\right)^{\mathrm{2}} }{\mathrm{2}\left({x}−\mathrm{2}\right)}−\int\frac{\left[{ln}\left(\left({x}−\mathrm{2}\right)+\mathrm{2}\right)\right]^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${x}−\mathrm{2}={t}\:\:\:\:{dx}={dt} \\ $$$$−\int\frac{\left[{ln}\left({t}+\mathrm{2}\right)\right]^{\mathrm{2}} }{{t}^{\mathrm{2}} }{dt} \\ $$$${u}^{'} =−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightarrow\:\:{u}=\frac{\mathrm{1}}{{t}} \\ $$$${v}=\left[{ln}\left({t}+\mathrm{2}\right)\right]^{\mathrm{2}} \rightarrow{v}'=\mathrm{2}{ln}\left({t}+\mathrm{2}\right)\frac{\mathrm{1}}{{t}+\mathrm{2}} \\ $$$${uv}=\frac{{ln}\left({t}+\mathrm{2}\right)}{{t}}\:\:\:{uv}'=\frac{\mathrm{2}{ln}\left({t}+\mathrm{2}\right)}{{t}} \\ $$$$=\mathrm{2}\frac{{lnx}}{{x}−\mathrm{2}} \\ $$$${I}=\frac{{x}\left({lnx}\right)^{\mathrm{2}} }{\mathrm{2}\left({x}−\mathrm{2}\right)}+\frac{{lnx}}{{x}−\mathrm{2}}−\mathrm{2}\int\frac{{lnx}}{{x}−\mathrm{2}} \\ $$$$\mathrm{3}{I}=\left[\frac{{x}\left({lnx}\right)^{\mathrm{2}} }{\mathrm{2}\left({x}−\mathrm{2}\right)}\right]_{\frac{\mathrm{3}}{\mathrm{2}}} ^{\mathrm{1}} +\left[\frac{{lnx}}{{x}−\mathrm{2}}\right]_{\frac{\mathrm{3}}{\mathrm{2}}} ^{\mathrm{1}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}\left[\left(−\frac{\frac{\mathrm{3}}{\mathrm{2}}\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{2}\right)}\right)−\frac{{ln}\mathrm{3}−{ln}\mathrm{2}}{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{3}\left(\right.}{}\frac{\left.{ln}\mathrm{3}−{ln}\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+\mathrm{2}×\frac{{ln}\mathrm{3}−{ln}\mathrm{2}}{\mathrm{3}} \\ $$$$=\frac{\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{2}\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)}{\mathrm{3}} \\ $$$${I}=\frac{\mathrm{3}\left({ln}\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\left({ln}\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)}{\mathrm{6}} \\ $$$$ \\ $$