Question

The value of ${\int }_0^{\infty }\frac{1}{\left[\left(x^2+4\right)\left(x^2+9\right)\right]}dx$ is

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

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{60}$

The explanation for the correct answer :

Solve for the given integrals.

$$\begin{gathered} I={\int }_0^{\infty }\frac{1}{\left[\left(x^2+4\right)\left(x^2+9\right)\right]}dx \\ =\frac{1}{5}\left[{\int }_0^{\infty }\frac{1}{\left[\left(x^2+4\right)\right]}dx-{\int }_0^{\infty }\frac{1}{\left[\left(x^2+9\right)\right]}dx\right] \\ =\frac{1}{5}\left\{{\left[\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right]}_0^{\infty }-{\left[\frac{1}{3}{\tan }^{-1}\frac{x}{3}\right]}_0^{\infty }\right\} \\ =\frac{1}{5}\left[\frac{1}{2}\times \frac{\mathrm{\pi }}{2}-0-\left(\frac{1}{3}\times \frac{\mathrm{\pi }}{2}-0\right)\right] \\ =\frac{1}{5}\left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{6}\right) \\ =\frac{\mathrm{\pi }}{60} \end{gathered}$$[$\therefore$$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$]

Hence, option (A) is the correct answer.