Integral Calculus Question 487
Question: $ \int_{{}}^{{}}{\frac{dx}{4x^{2}+9}=} $
[MP PET 1991; Roorkee 1977; MNR 1974]
Options:
A) $ \frac{1}{2}{{\tan }^{-1}}( \frac{2x}{3} )+c $
B) $ \frac{3}{2}{{\tan }^{-1}}( \frac{2x}{3} )+c $
C) $ \frac{1}{6}{{\tan }^{-1}}( \frac{2x}{3} )+c $
D) $ \frac{1}{6}{{\tan }^{-1}}( \frac{3x}{2} )+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{dx}{4x^{2}+9}=\frac{1}{4}\int_{{}}^{{}}{\frac{dx}{x^{2}+{{(3/2)}^{2}}}}} $ $ =\frac{1}{4}.\frac{2}{3}.{{\tan }^{-1}}( \frac{2x}{3} )+c=\frac{1}{6}{{\tan }^{-1}}( \frac{2x}{3} )+c. $
BETA
