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