Đặt: $\sqrt{1+\ln x}=t \Rightarrow 1+\ln x=t^2 \Rightarrow \dfrac{dx}{x}=2tdt$
Đổi cận: $x=1 \Rightarrow t=1$
$x=e \Rightarrow t=\sqrt{2}$
Ta có:
$I=\int\limits_1^{\sqrt 2}\dfrac{t^2-1}{t}2tdt$
$=2\int\limits_1^{\sqrt 2}(t^2-1)dt$
$=\left(\dfrac{2t^3}{3}-2t\right)\left|\begin{array}{l}\sqrt2\\1\end{array}\right.=\dfrac{4-2\sqrt2}{2}$