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

Solution

1. Use partial fraction decomposition to rewrite the integrand: $\frac{1}{(x^2+a^2)(x^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{1}{x^2+a^2}-\frac{1}{x^2+b^2}\right)$.
2. Integrate the expression from $0$ to $\infty$: $\frac{1}{b^2-a^2}{\left[\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)-\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right]}_0^{\infty }$.
3. Evaluate the limits to obtain $\frac{1}{b^2-a^2}\left(\frac{\pi }{2a}-\frac{\pi }{2b}\right)=\frac{\pi }{2(b^2-a^2)}\cdot \frac{b-a}{ab}$.
4. Simplify the expression using the difference of squares $b^2-a^2=(b-a)(b+a)$ to get $\frac{\pi }{2ab(a+b)}$.
5. Hence the answer is (A).