Question

Evaluate the following integral ${\int }_0^{\infty }\frac{dx}{(x^2+a^2)(x^2+b^2)}=$

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

Answer 1

The correct option is C $\frac{\pi }{2ab(a+b)}$

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

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

$\Rightarrow \frac{1}{(a^2-b^2)}{\int }_0^{\infty }\left[\frac{(x^2+a^2)-(x^2+b^2)}{(x^2+a^2)(x^2+b^2)}\right]dx$

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

$=\frac{1}{(a^2-b^2)}{[\frac{1}{b}ta{n}^{-1}\frac{x}{b}-\frac{1}{a}ta{n}^{-1}\frac{x}{a}]}_0^{\infty }$

$=\frac{1}{(a^2-b^2)}[\frac{1}{b}ta{n}^{-1}\infty -\frac{1}{a}ta{n}^{-1}\infty ]-(0-0)]$

$=\frac{1}{(a^2-b^2)}[\frac{1}{b}\times \frac{\pi }{2}-\frac{1}{a}\times \frac{\pi }{2}]$

$=\frac{\pi }{2(a-b)(a+b)}\left[\frac{(a-b)}{ab}\right]$

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