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