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

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(D) Let $I = \int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)}$.Using partial fractions,we can write $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{

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$\frac{\pi }{{2(a + b)}}$

Vedclass · Answer

(D) Let $I = \int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)}$.Using partial fractions,we can write $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}$.Solving for $A$ and $B$,we get $x^2 = A(x^2 + b^2) + B(x^2 + a^2)$.Setting $x^2 = -a^2$,we get $-a^2 = A(b^2 - a^2)$,so $A = \frac{a^2}{a^2 - b^2}$.Setting $x^2 = -b^2$,we get $-b^2 = B(a^2 - b^2)$,so $B = \frac{-b^2}{a^2 - b^2}$.Thus,$I = \frac{1}{a^2 - b^2} \int_0^\infty \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} ight) dx$.$I = \frac{1}{a^2 - b^2} \left[ a^2 \cdot \frac{1}{a} 	an^{-1}(\frac{x}{a}) - b^2 \cdot \frac{1}{b} 	an^{-1}(\frac{x}{b}) ight]_0^\infty$.$I = \frac{1}{a^2 - b^2} \left[ a \cdot \frac{\pi}{2} - b \cdot \frac{\pi}{2} ight] = \frac{\pi}{2(a^2 - b^2)} (a - b) = \frac{\pi}{2(a + b)}$.

$\int_0^\infty {\frac{{{x^2}\,dx}}{{({x^2} + {a^2})({x^2} + {b^2})}}} = $

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