Question

Given that $\underset{0}{\overset{\infty }{\int }}\frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)\left(x^2+c^2\right)}dx$
= $\frac{\pi }{2\left(a+b\right)\left(b+c\right){\left(c+a\right)}^{\prime }}$ then
$\underset{0}{\overset{\infty }{\int }}\frac{dx}{\left(x^2+4\right)\left(x^2+9\right)}$ is

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

Answer 1

Answer: (A)

Let
$I=\underset{0}{\overset{\infty }{\int }}\frac{dx}{\left(x^2+4\right)\left(x^2+9\right)}$
$=\underset{0}{\overset{\infty }{\int }}\frac{x^2}{x^2\left(x^2+4\right)\left(x^2+9\right)}dx$
$=\underset{0}{\overset{\infty }{\int }}\frac{x^2}{\left(x^2+0\right)\left(x^2+4\right)\left(x^2+9\right)}dx$
$=\frac{\pi }{2(0+2)(2+3)(3+0)}$
$[\because \underset{0}{\overset{\infty }{\int }}\frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)\left(x^2+c^2\right)}dx=\frac{\pi }{2(a+b)(b+c)(c+a)}]$
$=\frac{\pi }{60}$