Evaluate: `int_0^oo1/((x^2+a^2)(x^2+b^2))dxdot`

Explanation

Given

`int_0^oo1/((x^2+a^2)(x^2+b^2))dx`

`I=int_0^oo1/((x^2+a^2)(x^2+b^2))dx`

`I=int_0^oo[(x^2+a^2)-(x^2+b^2)]/((x^2+a^2)(x^2+b^2))dx`

On simplifying we get

`I=1/(a^2-b^2)int_0^oo (dx)/(x^2+b^2)-1/(a^2-b^2)int_0^oo (dx)/(x^2+a^2)`

`I=1/(a^2-b^2)\times(1/b)[tan^(-1)(x/b)]_0^oo-1/(a^2-b^2)\times(1/a)[tan^(-1)(x/a)]_0^oo`

`implies I=(pi)/(2ab(a+b))`