Solveeit Logo
Login

Question

Question: \[\int_{0}^{\infty}\frac{x^{2}dx}{(x^{2} + a^{2})(x^{2} + b^{2})} =\]...

∫0∞x2dx(x2+a2)(x2+b2)=\int_{0}^{\infty}\frac{x^{2}dx}{(x^{2} + a^{2})(x^{2} + b^{2})} =

A

π2(a−b)\frac{\pi}{2(a - b)}

B

π2(b−a)\frac{\pi}{2(b - a)}

C

π(a+b)\frac{\pi}{(a + b)}

D

π2(a+b)\frac{\pi}{2(a + b)}

Answer

π2(a+b)\frac{\pi}{2(a + b)}

Explanation

Solution

∫0∞x2dx(x2+a2)(x2+b2)=∫0∞(x2+a2)−a2(x2+a2)(x2+b2) dx\int_{0}^{\infty}{\frac{x^{2}dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \int_{0}^{\infty}{\frac{(x^{2} + a^{2}) - a^{2}}{(x^{2} + a^{2})(x^{2} + b^{2})}\ }dx}

∫0∞1x2+b2dx−a2∫0∞1(x2+a2)(x2+b2) dx=[1btan⁡−1xb]0∞−a2(a2−b2)∫0∞(1x2+b2−1x2+a2) dx\int_{0}^{\infty}{\frac{1}{x^{2} + b^{2}}dx - a^{2}\int_{0}^{\infty}{\frac{1}{(x^{2} + a^{2})(x^{2} + b^{2})}\ }dx} = \left\lbrack \frac{1}{b}\tan^{- 1}\frac{x}{b} \right\rbrack_{0}^{\infty} - \frac{a^{2}}{(a^{2} - b^{2})}\int_{0}^{\infty}{\left( \frac{1}{x^{2} + b^{2}} - \frac{1}{x^{2} + a^{2}} \right)\ }dx

=1b.π2−a2(a2−b2)[1btan⁡−1xb−1atan⁡−1xa]0∞=π2(a+b)= \frac{1}{b}.\frac{\pi}{2} - \frac{a^{2}}{(a^{2} - b^{2})}\left\lbrack \frac{1}{b}\tan^{- 1}\frac{x}{b} - \frac{1}{a}\tan^{- 1}\frac{x}{a} \right\rbrack_{0}^{\infty} = \frac{\pi}{2(a + b)}.