Đáp án:
\[I = \frac{3}{4}{e^2} + \frac{7}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_1^e {\left( {3x + 1} \right)\ln xdx} \\
\left\{ \begin{array}{l}
u = \ln x\\
v' = 3x + 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = \frac{{3{x^2}}}{2} + x
\end{array} \right.\\
\Rightarrow I = \int\limits_1^e {\left( {3x + 1} \right)\ln xdx} \\
= \mathop {\left. {\ln x.\left( {\frac{{3{x^2}}}{2} + x} \right)} \right|}\nolimits_1^e - \int\limits_1^e {\frac{1}{x}.\left( {\frac{3}{2}{x^2} + x} \right)dx} \\
= \ln e.\left( {\frac{{3{e^2}}}{2} + e} \right) - \ln 1.\left( {\frac{3}{2} + 1} \right) - \int\limits_1^e {\left( {\frac{3}{2}x + 1} \right)dx} \\
= \left( {\frac{3}{2}{e^2} + e} \right) - \mathop {\left. {\left( {\frac{3}{4}{x^2} + x} \right)} \right|}\nolimits_1^e \\
= \left( {\frac{3}{2}{e^2} + e} \right) - \left( {\frac{3}{4}{e^2} + e} \right) + \left( {\frac{3}{4}{{.1}^2} + 1} \right)\\
= \frac{3}{4}{e^2} + \frac{7}{4}
\end{array}\)
