Đặt $t = {e^x} \Leftrightarrow dt = {e^x}dx$$\begin{cases} x = \ln 2 \Rightarrow t = 2 \\ x = 0 \Rightarrow t = 1 \end{cases}$
$I = \int\limits_1^2 {\frac{{dt}}{{{t^2} - 9}}} = \int\limits_1^2 {\frac{{dt}}{{(t - 3)(t + 3)}}} = \frac{1}{6}\int\limits_1^2 {\left( {\frac{1}{{t - 3}} - \frac{1}{{t + 3}}} \right)} dt = \frac{1}{6}\left( {\ln \left| {t - 3} \right| - \ln \left| {t + 3} \right|} \right)\begin{cases}2 \\ 1 \end{cases} = \frac{1}{6}\ln \frac{2}{5}$