Đặt: $t=1+\ln x \Rightarrow dt=\dfrac{dx}{x}$
Đổi cận: $x=1 \Rightarrow t=1$
$x=e \Rightarrow t=2$
Ta có:
$I=\int\limits_1^2\dfrac{t-1+\sqrt t}{\sqrt t}dt$
$=\int\limits_1^2\left(\sqrt t-1-\dfrac{1}{\sqrt t}\right)dx$
$=\left(\dfrac{2}{3}t\sqrt t-t-2\sqrt t\right)\left|\begin{array}{l}2\\1\end{array}\right.$
$=\dfrac{7-2\sqrt2}{3}$