Đáp án:
\[I = \frac{{14}}{9}\]
Giải thích các bước giải:
\(\begin{array}{l}
I = \int\limits_1^e {\frac{{\sqrt {3\ln x + 1} }}{x}dx} = \int\limits_1^e {\sqrt {3\ln x + 1} .\frac{{dx}}{x}} \\
t = \sqrt {3\ln x + 1} \Rightarrow {t^2} = 3\ln x + 1\\
\Leftrightarrow \left( {{t^2}} \right)'dt = \left( {3\ln x + 1} \right)'dx\\
\Leftrightarrow 2tdt = \frac{3}{x}dx\\
\Leftrightarrow \frac{{dx}}{x} = \frac{2}{3}tdt\\
\left\{ \begin{array}{l}
x = 1 \Rightarrow t = \sqrt {3.\ln 1 + 1} = 1\\
x = e \Rightarrow t = \sqrt {3\ln e + 1} = 2
\end{array} \right.\\
\Rightarrow I = \int\limits_1^e {\sqrt {3\ln x + 1} .\frac{{dx}}{x}} \\
= \int\limits_1^2 {t.\frac{2}{3}tdt} \\
= \frac{2}{3}\int\limits_1^2 {{t^2}dt} \\
= \mathop {\left. {\frac{2}{3}.\frac{1}{3}{t^3}} \right|}\nolimits_1^2 \\
= \frac{2}{9}\left( {{2^3} - {1^3}} \right)\\
= \frac{{14}}{9}
\end{array}\)
