Mathematics - Definite Integration Question with Solution | TestHub

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