Definite-Integration Question 510
Question: $ \int_0^{\infty }{\frac{x^{2},dx}{(x^{2}+a^{2})(x^{2}+b^{2})}}= $
Options:
A) $ \frac{\pi }{2(a-b)} $
B) $ \frac{\pi }{2(b-a)} $
C) $ \frac{\pi }{(a+b)} $
D) $ \frac{\pi }{2(a+b)} $
Show Answer
Answer:
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})}}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})}}dx} $ $ =[ \frac{1}{b}{{\tan }^{-1}}\frac{x}{b} ]_0^{\infty }-\frac{a^{2}}{(a^{2}-b^{2})}\int_0^{\infty }{( \frac{1}{x^{2}+b^{2}}-\frac{1}{x^{2}+a^{2}} )}dx $ $ =\frac{1}{b}.\frac{\pi }{2}-\frac{a^{2}}{(a^{2}-b^{2})}[ \frac{1}{b}{{\tan }^{-1}}\frac{x}{b}-\frac{1}{a}{{\tan }^{-1}}\frac{x}{a} ]_0^{\infty }=\frac{\pi }{2(a+b)} $ .
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