Đáp án:
\[\int\limits_1^e {\left( {2x + 1} \right)\ln x} = \frac{1}{2}{e^2} + \frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = \ln x\\
v' = 2x + 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = {x^2} + x
\end{array} \right.\\
\Rightarrow \int\limits_1^e {\left( {2x + 1} \right)\ln x} = \mathop {\left. {\left( {{x^2} + x} \right)\ln x} \right|}\nolimits_1^e - \int\limits_1^e {\frac{1}{x}.\left( {{x^2} + x} \right)dx} \\
= \left( {{e^2} + e} \right)\ln e - \left( {{1^2} + 1} \right).\ln 1 - \int\limits_1^e {\left( {x + 1} \right)dx} \\
= \left( {{e^2} + e} \right) - \mathop {\left. {\left( {\frac{{{x^2}}}{2} + x} \right)} \right|}\nolimits_1^e \\
= {e^2} + e - \left( {\frac{{{e^2}}}{2} + e} \right) + \left( {\frac{{{1^2}}}{2} + 1} \right)\\
= \frac{1}{2}{e^2} + \frac{3}{2}
\end{array}\)
