Ta có $I = \int\limits_0^{3\ln 2} {\frac{{{e^{\frac{x}{3}}}dx}}{{{e^{\frac{x}{3}}}{{({e^{\frac{x}{3}}} + 2)}^2}}}} $
Đặt $u=e^{\frac{x}{3}}\Rightarrow $$3du = {e^{\frac{x}{3}}}dx$;
Đổi cận $x = 0 \Rightarrow u = 1;x = 3\ln 2 \Rightarrow u = 2$
Ta được $I = \int\limits_1^2 {\frac{{3du}}{{u{{(u + 2)}^2}}}} =3$$\int\limits_1^2 {\left( {\frac{1}{{4u}} - \frac{1}{{4(u + 2)}}} \right.\left. { - \frac{1}{{2{{(u + 2)}^2}}}} \right)} du$
                    $=3\left. {\left( {\frac{1}{4}\ln \left| u \right| - \frac{1}{4}\ln \left| {u + 2} \right| + \frac{1}{{2(u + 2)}}} \right)} \right|_1^2$
                    $\boxed{= \frac{3}{4}\ln (\frac{3}{2}) - \frac{1}{8}} $

Đặt
$t = \pi  - x \Rightarrow dt =  - dx$
Đổi cận $\begin{cases} x=0 \rightarrow t=\pi \\ x=\pi \rightarrow t=0 \end{cases}$

$\begin{array}{l}  \Rightarrow I =- \int\limits_\pi^0  {\frac{{\left( {\pi  - t} \right)\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt= \int\limits_0^\pi  {\frac{{\left( {\pi  - t} \right)\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt = \pi \int\limits_0^\pi  {\frac{{\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt - I\\
 \Rightarrow 2I = \pi \int\limits_0^\pi  {\frac{{\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt =  - \pi \int\limits_0^\pi  {\frac{{d(\cos t)}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}}  = -\pi \arctan \cos t|_0^\pi=\pi \left( {\frac{\pi }{4} +\frac{\pi }{4}} \right)\\
 \Rightarrow \boxed{I =\frac{\pi^2}{4} }
\end{array}$