Integral Calculus Question 228
Question: $ \int\limits_0^{1}{\frac{1}{( x^{2}+16 )( x^{2}+25 )}dx=} $
Options:
A) $ \frac{1}{5}[ \frac{1}{4}{{\tan }^{-1}}( \frac{1}{4} )-\frac{1}{5}{{\tan }^{-1}}( \frac{1}{5} ) ] $
B) $ \frac{1}{9}[ \frac{1}{4}{{\tan }^{-1}}( \frac{1}{4} )-\frac{1}{5}{{\tan }^{-1}}( \frac{1}{5} ) ] $
C) $ \frac{1}{4}[ \frac{1}{4}{{\tan }^{-1}}( \frac{1}{4} )-\frac{1}{5}{{\tan }^{-1}}( \frac{1}{5} ) ] $
D) $ \frac{1}{9}[ \frac{1}{5}{{\tan }^{-1}}( \frac{1}{4} )-\frac{1}{5}{{\tan }^{-1}}( \frac{1}{5} ) ] $
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Answer:
Correct Answer: B
Solution:
[b] Let $ I=\int\limits_0^{1}{\frac{dx}{(x^{2}+16)(x^{2}+25)}} $ $ =\frac{1}{9}\int\limits_0^{1}{( \frac{1}{x^{2}+16}-\frac{1}{x^{2}+25} )dx} $ $ =\frac{1}{9}( \frac{1}{4}{{\tan }^{-1}}\frac{x}{4}-\frac{1}{5}{{\tan }^{-1}}\frac{x}{5} )_0^{1} $ $ =\frac{1}{9}[ \frac{1}{4}{{\tan }^{-1}}\frac{1}{4}-\frac{1}{5}{{\tan }^{-1}}\frac{1}{5} ] $
BETA
