Question

The value of ${\int }_0^{\infty }\frac{dx}{(x^2+4)(x^2+9)}$ is

• A. $\frac{\pi }{60}$
• B. $\frac{\pi }{20}$
• C. $\frac{\pi }{40}$
• D. $\frac{\pi }{80}$

Answer 1

The correct option is A $\frac{\pi }{60}$

Let $I={\int }_0^{\infty }\frac{dx}{(x^2+4)(x^2+9)}$
$=\frac{1}{5}\left({\int }_0^{\infty }\frac{1}{(x^2+4)}dx-{\int }_0^{\infty }\frac{1}{(x^2+9)}dx\right)$
$=\frac{1}{5}[{\left[\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right]}_0^{\infty }-{\left[\frac{1}{3}{\tan }^{-1}\frac{x}{3}\right]}_0^{\infty }]$
$=\frac{1}{5}[\frac{1}{2}\cdot \frac{\pi }{2}-0-(\frac{1}{3}\cdot \frac{\pi }{2}-0)]$
$=\frac{1}{5}(\frac{\pi }{4}-\frac{\pi }{6})=\frac{\pi }{60}$