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