Integral Calculus Question 52
Question: $ \int_{{}}^{{}}{\frac{1}{(x^{2}+a^{2})(x^{2}+b^{2})}dx=} $
Options:
A) $ \frac{1}{(a^{2}-b^{2})}[ \frac{1}{b}{{\tan }^{-1}}( \frac{x}{b} )-\frac{1}{a}{{\tan }^{-1}}( \frac{x}{a} ) ]+c $
B) $ \frac{1}{(b^{2}-a^{2})}[ \frac{1}{b}{{\tan }^{-1}}( \frac{x}{b} )-\frac{1}{a}{{\tan }^{-1}}( \frac{x}{a} ) ]+c $
C) $ \frac{1}{b}{{\tan }^{-1}}( \frac{x}{b} )-\frac{1}{a}{{\tan }^{-1}}( \frac{x}{a} )+c $
D) $ \frac{1}{a}{{\tan }^{-1}}( \frac{x}{a} )-\frac{1}{b}{{\tan }^{-1}}( \frac{x}{b} )+c $
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Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{1}{(x^{2}+b^{2})(x^{2}+a^{2})}},dx $ $ =\frac{1}{a^{2}-b^{2}}\int_{{}}^{{}}{[ \frac{1}{x^{2}+b^{2}}-\frac{1}{x^{2}+a^{2}} ]},dx $ $ =\frac{1}{(a^{2}-b^{2})}[ \frac{1}{b}{{\tan }^{-1}}( \frac{x}{b} )-\frac{1}{a}{{\tan }^{-1}}( \frac{x}{a} ) ]+c $ . Note : Students should remember this question as a formula.
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