Đáp án:
\[I = \frac{{{e^2} + 1}}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = \ln x\\
v' = x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = \frac{{{x^2}}}{2}
\end{array} \right.\\
\Rightarrow I = \int\limits_1^e {x.\ln x.dx} \\
= \mathop {\left. {\frac{{{x^2}}}{2}.\ln x} \right|}\nolimits_1^e - \int\limits_1^e {\frac{1}{x}.\frac{{{x^2}}}{2}dx} \\
= \frac{{{e^2}}}{2} - \frac{1}{2}\int\limits_1^e {xdx} \\
= \frac{{{e^2}}}{2} - \mathop {\left. {\frac{1}{4}{x^2}} \right|}\nolimits_1^e \\
= \frac{{{e^2}}}{2} - \frac{1}{4}\left( {{e^2} - 1} \right)\\
= \frac{{{e^2} + 1}}{4}
\end{array}\)
