A) \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{2x}{3} \right)+c\]
B) \[\frac{3}{2}{{\tan }^{-1}}\left( \frac{2x}{3} \right)+c\]
C) \[\frac{1}{6}{{\tan }^{-1}}\left( \frac{2x}{3} \right)+c\]
D) \[\frac{1}{6}{{\tan }^{-1}}\left( \frac{3x}{2} \right)+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{dx}{4{{x}^{2}}+9}=\frac{1}{4}\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}+{{(3/2)}^{2}}}}}\] \[=\frac{1}{4}.\frac{2}{3}.{{\tan }^{-1}}\left( \frac{2x}{3} \right)+c=\frac{1}{6}{{\tan }^{-1}}\left( \frac{2x}{3} \right)+c.\]
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