STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$= \int_0^\infty {\frac{1}{{{x^2} + {b^2}}}dx - {a^2}\int_0^\infty {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}{\rm{ }}} dx} $

$ = \left[ {\frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b}} \right]_0^\infty - \frac{{{a^2}}}{{({a^2} - {b^2})}}\int_0^\infty {\left( {\frac{1}{{{x^2} + {b^2}}} - \frac{1}{{{x^2} + {a^2}}}} \right){\rm{ }}} dx$

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

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