$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=$

Question

Aicharya · Accepted Answer

Let $x^{2}=t$ $\begin{aligned} 	herefore & \frac{1}{\left(x^{2}+4ight)\left(x^{2}+9ight)}=\frac{1}{(t+4)(t+9)} 	ext { and let } \frac{1}{(t+4)(t+9)}=\frac{1}{5}\left[\frac{1}{t+4}-\frac{1}{t+9}ight] \ 	herefore I &=\frac{1}{5} \int_{0}^{\infty}\left[\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}ight] \mathrm{dx} \ &=\frac{1}{5}\left\{\left[\frac{1}{2} 	an ^{-1}\left(\frac{x}{2}ight)ight]_{0}^{\infty}-\left[\frac{1}{3} 	an ^{-1}\left(\frac{x}{3}ight)ight]_{0}^{\infty}ight\} \ &=\frac{1}{5}\left\{\frac{1}{2} 	imes \frac{\pi}{2}-\frac{1}{3} 	imes \frac{\pi}{2}ight\}=\frac{\pi}{10}\left\{\frac{1}{2}-\frac{1}{3}ight\}=\frac{\pi}{60} \end{aligned}$

$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=$

Solution