Đặt: $t=e^x+1 \Rightarrow dt=e^xdx$
Đổi cận: $x=\ln3 \Rightarrow t=4$
$x=\ln 8 \Rightarrow t=9$
Ta có:
$I=\int\limits_4^9\dfrac{\ln(t-1)}{\sqrt t}dt$
$=2\int\limits_4^9\ln(t-1)d(\sqrt t)$
$=2\ln(t-1)\sqrt t\left|\begin{array}{l}9\\4\end{array}\right.-2\int\limits_4^9\sqrt t d(\ln(t-1))$
$=6\ln8-4\ln3-2\int\limits_4^9\dfrac{\sqrt tdt}{t-1}$
$=6\ln8-4\ln3-2\int\limits_4^9\dfrac{2td(\sqrt t)}{t-1}$
$=6\ln8-4\ln3-2\int\limits_4^9\left(2+\dfrac{1}{\sqrt t-1}-\dfrac{1}{\sqrt t+1}\right)d(\sqrt t)$
$=6\ln8-4\ln3-2(2\sqrt t+\ln(\sqrt t-1)-\ln(\sqrt t+1))\left|\begin{array}{l}9\\4\end{array}\right.$
$=6\ln8-4\ln3-4+\ln\dfrac{4}{9}$