Let $x^{2}=t$
$\begin{aligned}
\therefore & \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\frac{1}{(t+4)(t+9)} \text { and let } \frac{1}{(t+4)(t+9)}=\frac{1}{5}\left[\frac{1}{t+4}-\frac{1}{t+9}\right] \\
\therefore I &=\frac{1}{5} \int_{0}^{\infty}\left[\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right] \mathrm{dx} \\
&=\frac{1}{5}\left\{\left[\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]_{0}^{\infty}-\left[\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{\infty}\right\} \\
&=\frac{1}{5}\left\{\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{3} \times \frac{\pi}{2}\right\}=\frac{\pi}{10}\left\{\frac{1}{2}-\frac{1}{3}\right\}=\frac{\pi}{60}
\end{aligned}$