\[\int_{{}}^{{}}{\frac{dx}{({{x}^{2}}+1)({{x}^{2}}+4)}=}\] [MP PET 1995]

\[{{\tan }^{-1}}x-2{{\tan }^{-1}}\frac{x}{2}+c\]

\[\frac{1}{3}{{\tan }^{-1}}x-\frac{1}{3}{{\tan }^{-1}}\frac{x}{2}+c\]

\[\frac{1}{3}{{\tan }^{-1}}x+\frac{1}{3}{{\tan }^{-1}}\frac{x}{2}+c\]

\[\frac{1}{3}{{\tan }^{-1}}x-\frac{1}{6}{{\tan }^{-1}}\frac{x}{2}+c\]

\[\int_{{}}^{{}}{\frac{dx}{({{x}^{2}}+1)({{x}^{2}}+4)}=}\]

1.	\[\int_{{}}^{{}}{\frac{dx}{({{x}^{2}}+1)({{x}^{2}}+4)}=}\] [MP PET 1995]
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.	\[{{\tan }^{-1}}x-2{{\tan }^{-1}}\frac{x}{2}+c\]
Answer» D. \[{{\tan }^{-1}}x-2{{\tan }^{-1}}\frac{x}{2}+c\]

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\[\int_{{}}^{{}}{\frac{dx}{({{x}^{2}}+1)({{x}^{2}}+4)}=}\] [MP PET 1995]

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