Question

$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$ is

• A. $\frac{\pi a b}{a+b}$
• B. $\frac{\pi}{2(a+b)}$
• C. $\frac{\pi}{2 ab ( a + b )}$
• D. $\frac{\pi(a+b)}{2 a b}$

Answer 1

Answer: (C)

$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$
$=\frac{1}{b^{2}-a^{2}} \int_{0}^{\infty} \frac{\left(x^{2}+b^{2}\right)-\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$
$=\frac{1}{b^{2}-a^{2}} \int_{0}^{\infty}\left[\frac{1}{x^{2}+a^{2}}-\frac{1}{x^{2}+b^{2}}\right] d x$
$=\frac{1}{b^{2}-a^{2}}\left[\frac{1}{a} \tan ^{-1} \frac{x}{a}-\frac{1}{b} \tan ^{-1} \frac{x}{b}\right]_{0}^{\infty}$
$=\frac{1}{b^{2}-a^{2}}\left[\frac{\pi}{2 a}-\frac{\pi}{2 b}\right]=\frac{\pi}{2 a b(a+b)}$