The value of int_0^{infty} frac{dx}{(x^2+4)(x | vedclass

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(A) Let $I = \int_0^{\infty} \frac{dx}{(x^2+4)(x^2+9)}$.Using partial fractions, we can write $\frac{1}{(x^2+4)(x^2+9)} = \frac{1}{5} \left( \frac{1}{

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$\frac{\pi}{60}$

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(A) Let $I = \int_0^{\infty} \frac{dx}{(x^2+4)(x^2+9)}$.Using partial fractions, we can write $\frac{1}{(x^2+4)(x^2+9)} = \frac{1}{5} \left( \frac{1}{x^2+4} - \frac{1}{x^2+9} ight)$.Therefore, $I = \frac{1}{5} \left[ \int_0^{\infty} \frac{dx}{x^2+2^2} - \int_0^{\infty} \frac{dx}{x^2+3^2} ight]$.Using the standard integral formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$, we get:$I = \frac{1}{5} \left[ \left( \frac{1}{2} 	an^{-1}(\frac{x}{2}) ight)_0^{\infty} - \left( \frac{1}{3} 	an^{-1}(\frac{x}{3}) ight)_0^{\infty} ight]$.Evaluating the limits:$I = \frac{1}{5} \left[ \frac{1}{2} (\frac{\pi}{2} - 0) - \frac{1}{3} (\frac{\pi}{2} - 0) ight]$.$I = \frac{1}{5} \left[ \frac{\pi}{4} - \frac{\pi}{6} ight] = \frac{1}{5} \left[ \frac{3\pi - 2\pi}{12} ight] = \frac{1}{5} \cdot \frac{\pi}{12} = \frac{\pi}{60}$.

The value of $\int_0^{\infty} \frac{dx}{(x^2+4)(x^2+9)}$ is

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