Question

Solve:
${\int }_0^{\infty }\frac{1}{(x^2+a^2)(x^2+b^2)}dx$

Answer 1

${\int }_0^{\infty }\frac{1}{(x^2+a^2)(x^2+b^2)}$

$\frac{1}{a^2-b^2}{\int }_0^{\infty }\frac{(x^2+a^2)-(x^2+b^2)}{(x^2+a^2)-(x^2+b^2)}$

$=\frac{1}{a^2-b^2}{\int }_0^{\infty }(\frac{1}{x^2+b^2}-\frac{1}{x^2+a^2})dx$

$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}$

$\int \frac{1}{x^2+b^2}dx=\frac{1}{b}{\tan }^{-1}\frac{x}{b}$

$=\frac{1}{a^2-b^2}{\left[\frac{1}{b}{\tan }^{-1}\frac{x}{b}\right]}_0^{\infty }-{\left[\frac{1}{a}{\tan }^{-1}\frac{x}{a}\right]}_0^{\infty }$

$=\frac{1}{a^2-b^2}\left[\frac{1}{b}\right({\tan }^{-1}\infty -{\tan }^{-1}0)-\frac{1}{a}][{\tan }^{-1}\infty -{\tan }^{-1}0]$

$=\frac{1}{a^2-b^2}|\frac{1}{b}\times \frac{\pi }{2}-\frac{\pi }{2}\times \frac{1}{a}=\frac{\pi }{2}|$

$\frac{\pi }{2}\times \frac{1}{a^2-b^2}|\frac{a-b}{ab}|=\frac{\pi }{2}=\frac{1}{a+b}\times \frac{1}{a-b}$.