Tính tích phân sau (I = intlimits_1^e {xln x,dx} ).

Đặt (left{ begin{array}{l}u = ln x\dv = xdxend{array} right. Rightarrow left{ begin{array}{l}du = dfrac{1}{x}dx\v = dfrac{{{x^2}}}{2}end{array} right.)

Khi đó ta có: (I = intlimits_1^e {xln x,dx}  )

(= left( {dfrac{{{x^2}}}{2}ln x} right)left| begin{array}{l}^e\_1end{array} right. – intlimits_1^e {dfrac{x}{2}} ,dx )

(= dfrac{{{e^2}}}{2} – left( {dfrac{{{x^2}}}{4}} right)left| begin{array}{l}^e\_1end{array} right. )

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

Chọn đáp án C.