Given that int_0^infty frac{x^2 , dx}{(x^2 + | vedclass

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(A) The given formula is $\int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)(x^2 + c^2)} = \frac{\pi}{2(a + b)(b + c)(c + a)}$.To find the value of

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$\frac{\pi}{60}$

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(A) The given formula is $\int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)(x^2 + c^2)} = \frac{\pi}{2(a + b)(b + c)(c + a)}$.To find the value of $I = \int_0^\infty \frac{x^2 \, dx}{(x^2 + 4)(x^2 + 9)}$,we can rewrite the denominator as $(x^2 + 2^2)(x^2 + 3^2)(x^2 + 0^2)$.Comparing this with the given formula,we have $a = 2$,$b = 3$,and $c = 0$.Substituting these values into the right-hand side of the formula:$I = \frac{\pi}{2(2 + 3)(3 + 0)(0 + 2)}$$I = \frac{\pi}{2(5)(3)(2)}$$I = \frac{\pi}{60}$.

Given that $\int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)(x^2 + c^2)} = \frac{\pi}{2(a + b)(b + c)(c + a)}$,then the value of $\int_0^\infty \frac{x^2 \, dx}{(x^2 + 4)(x^2 + 9)}$ is

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