Đáp án:
$\begin{array}{l}
I = \int\limits_1^e {\frac{{{x^2} + 2{\mathop{\rm lnx}\nolimits} }}{x}dx} \\
= \int\limits_1^e {x + 2.\ln x.\frac{1}{x}dx} \\
= \left( {\frac{{{x^2}}}{2}} \right)_1^e + A\left( {A = 2\int\limits_1^e {\ln x.\frac{1}{x}dx} } \right)\\
= \frac{{{e^2} - 1}}{2} + A\\
Có:A = \left( {2.\ln x.\ln x} \right)_1^e - 2\int\limits_1^e {\ln x.\frac{1}{x}dx} \\
\Rightarrow A = 2 - A\\
\Rightarrow A = 1\\
\Rightarrow I = \frac{{{e^2} - 1}}{2} + 1 = \frac{{{e^2} + 1}}{2}
\end{array}$
