Tích phân sau (I = intlimits_1^e {2xleft( {1 – ln x} right),dx} ) bằng :

Ta có: (I = intlimits_1^e {2xleft( {1 – ln x} right),dx} )(, = intlimits_1^e {2x,dx}  – 2intlimits_1^e {xln ,dx})(,  = {x^2}left| {_1^e} right. – 2intlimits_1^e {xln ,dx} )

Đặt ({I_1} = intlimits_1^e {xln x,dx} )     

Ta có:

({I_1} = intlimits_1^e {xln x,dx}  = left( {dfrac{{{x^2}}}{2}ln x} right)left| begin{array}{l}^e\_1^{}end{array} right. – intlimits_1^e {dfrac{x}{2}dx} )

(= left( {dfrac{{{x^2}}}{2}ln x} right)left| begin{array}{l}^e\_1^{}end{array} right. – left( {dfrac{{{x^2}}}{4}} right)left| begin{array}{l}_{}^e\_1^{}end{array} right.)

( = dfrac{e^2}{2}ln e – left( {dfrac{e^2}{4} – dfrac{1}{4}} right) = dfrac{e^2}{2}+dfrac {1}{4})

Khi đó ta có: (I = {e^2} – 1 – 2.left( {dfrac{{{e^2}}}{4} + dfrac{1}{4}} right) = dfrac{{{e^2} – 3}}{2})