Đặt $t=\sqrt{1+3\ln x}\Rightarrow t^2=1+3\ln x\Rightarrow 2tdt=\dfrac{3dx}{x}$
Đổi cận: $x=1 \Rightarrow t=1$
$x=e \Rightarrow t=2$
Suy ra:
$I=\int\limits_1^2\dfrac{t(t^2-1)}{3}.\dfrac{2tdt}{3}$
$=\dfrac{2}{9}\int\limits_1^2(t^4-t^2)dt$
$=\dfrac{2}{9}\left(\dfrac{t^5}{5}-\dfrac{t^3}{3}\right)\left|\begin{array}{l}2\\1\end{array}\right.$
$=\dfrac{116}{135}$