int_{0}^{infty} frac{dx}{(x^{2}+4)(x^{2}+9)} | vedclass

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Question

(B) Let $I = \int_{0}^{\infty} \frac{dx}{(x^{2}+4)(x^{2}+9)}$.Using partial fractions,we write $\frac{1}{(x^{2}+4)(x^{2}+9)} = \frac{A}{x^{2}+4} + \fr

Vedclass · Accepted Answer

$\frac{\pi}{60}$

Vedclass · Answer

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

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

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