Question

The value of $\underset{0}{\overset{\infty }{\int }}\frac{dx}{\left(x^2+4\right)\left(x^2+9\right)}$ is

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

Answer 1

Answer: (A)

Let $I=\underset{0}{\overset{\infty }{\int }}\frac{dx}{\left(x^2+4\right)\left(x^2+9\right)}$

$=\frac{1}{5}\left(\underset{0}{\overset{\infty }{\int }}\frac{1}{\left(x^2+4\right)}dx-\underset{0}{\overset{\infty }{\int }}\frac{1}{\left(x^2+9\right)}dx\right)$

$=\frac{1}{5}\left[{\left[\frac{1}{2}ta{n}^{-1}\frac{x}{2}\right]}_0^{\infty }-{\left[\frac{1}{3}ta{n}^{-1}\frac{x}{3}\right]}_0^{\infty }\right]$

$=\frac{1}{5}\left[\frac{1}{2}\cdot \frac{\pi }{2}-0-\left(\frac{1}{3}\cdot \frac{\pi }{2}-0\right)\right]$

$=\frac{1}{5}\left(\frac{\pi }{4}-\frac{\pi }{6}\right)$

$=\frac{\pi }{60}$