Đặt $t=\sqrt{e^{x}+1}\Rightarrow e^{x}=t^{2}-1\Rightarrow e^{x}dx=2tdt$
$x=\ln 3\Rightarrow t=2$
$x=\ln 8\Rightarrow t=3$
$I=\int\limits_{\ln 3}^{\ln 8}\dfrac{dx}{\sqrt{e^{x}+1}}=\int\limits_{\ln 3}^{\ln 8}\dfrac{e^{x}dx}{e^{x}\sqrt{e^{x}+1}}=\int\limits_{2}^{3}\dfrac{2tdt}{(t^{2}-1)t}$
$=2\int\limits_{2}^{3}\dfrac{dt}{(t-1)(t+1)}=2\int\limits_{2}^{3}\dfrac{1}{2}\left(\dfrac{1}{t+1}-\dfrac{1}{t-1}\right)dt=\left[\ln |t+1|-\ln |t-1|\right]\left|\begin{matrix}3 \\ 2\end{matrix}\right.$
$=\ln \left|\dfrac{t+1}{t-1}\right|\left| \begin{matrix}3 \\ 2\end{matrix}\right. =\ln \dfrac{2}{3}$