Integral Calculus Question 186
Question: $ \int\limits_0^{\infty }{\frac{dx}{(x^{2}+a^{2})(x^{2}+b^{2})}} $ is
Options:
A) $ \frac{\pi ab}{a+b} $
B) $ \frac{\pi }{2(a+b)} $
C) $ \frac{\pi }{2ab(a+b)} $
D) $ \frac{\pi (a+b)}{2ab} $
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Answer:
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 }{[ \frac{1}{x^{2}+a^{2}}-\frac{1}{x^{2}+b^{2}} ]dx} $ $ =\frac{1}{b^{2}-a^{2}}[ \frac{1}{a}{{\tan }^{-1}}\frac{x}{a}-\frac{1}{b}{{\tan }^{-1}}\frac{x}{b} ]_0^{\infty } $ $ =\frac{1}{b^{2}-a^{2}}[ \frac{\pi }{2a}-\frac{x}{2b} ]=\frac{\pi }{2ab(a+b)} $
BETA
