Question

If ${\int }_0^{\infty }\frac{x^{2}dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}=\frac{\pi }{2(a+b)(b+c)(c+a)}$, then the value of ${\int }_0^{\infty }\frac{1}{(x^2+4)(x^2+9)}dx$ is.

• A. $\frac{\pi }{360}$
• B. $\frac{\pi }{180}$
• C. $\frac{\pi }{40}$
• D. $\frac{\pi }{80}$

Answer 1

The correct option is A $\frac{\pi }{360}$

Given : ${\int }_0^{\infty }\frac{x^{2}dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}=\frac{\pi }{2(a+b)(b+c)(c+a)}$ $⟶\left(1\right)$

${\int }_0^{\infty }\frac{1}{(x^2+4)(x^2+9)}dx={\int }_0^{\infty }\frac{x^2}{(x^2+4)(x^2+9)x^2}dx$

$={\int }_0^{\infty }\frac{x^2}{(x^2+4)(x^2+9)(x^2+0)}dx$

$={\int }_0^{\infty }\frac{x^2}{(x^2+2^2)(x^2+3^2)(x^2+0^2)}dx$

comparing with eqn. (1)

${\int }_0^{\infty }\frac{1}{(x^2+4)(x^2+9)}dx=\frac{\pi }{2(2+3)(3+0)(0+2)}$

$=\frac{\pi }{2\times 5\times 3\times 2}$

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