A) \[\frac{\pi ab}{a+b}\]
B) \[\frac{\pi }{2(a+b)}\]
C) \[\frac{\pi }{2ab(a+b)}\]
D) \[\frac{\pi (a+b)}{2ab}\]
Correct Answer: C
Solution :
[c] \[\int\limits_{0}^{\infty }{\frac{dx}{({{x}^{2}}+{{a}^{2}})({{x}^{2}}+{{b}^{2}})}}\] \[=\frac{1}{{{b}^{2}}-{{a}^{2}}}\int\limits_{0}^{\infty }{\frac{({{x}^{2}}+{{b}^{2}})-({{x}^{2}}+{{a}^{2}})}{({{x}^{2}}+{{a}^{2}})({{x}^{2}}+{{b}^{2}})}}\] \[=\frac{1}{{{b}^{2}}-{{a}^{2}}}\int\limits_{0}^{\infty }{\left[ \frac{1}{{{x}^{2}}+{{a}^{2}}}-\frac{1}{{{x}^{2}}+{{b}^{2}}} \right]dx}\] \[=\frac{1}{{{b}^{2}}-{{a}^{2}}}\left[ \frac{1}{a}{{\tan }^{-1}}\frac{x}{a}-\frac{1}{b}{{\tan }^{-1}}\frac{x}{b} \right]_{0}^{\infty }\] \[=\frac{1}{{{b}^{2}}-{{a}^{2}}}\left[ \frac{\pi }{2a}-\frac{x}{2b} \right]=\frac{\pi }{2ab(a+b)}\]
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