\[\int{\frac{{{x}^{2}}}{({{x}^{2}}+1)({{x}^{2}}+4)}dx}\] is equal to

1.	\[\int{\frac{{{x}^{2}}}{({{x}^{2}}+1)({{x}^{2}}+4)}dx}\] is equal to
A.	\[{{\tan }^{-1}}x+2{{\tan }^{-1}}\left( \frac{x}{2} \right)+C\]
B.	\[{{\tan }^{-1}}\left( \frac{x}{2} \right)-4{{\tan }^{-1}}x+C\]
C.	\[-\frac{1}{3}{{\tan }^{-1}}x+\frac{2}{3}{{\tan }^{-1}}\left( \frac{x}{2} \right)+C\]
D.	\[4{{\tan }^{-1}}\left( \frac{x}{2} \right)-2{{\tan }^{-1}}x+C\]
Answer» D. \[4{{\tan }^{-1}}\left( \frac{x}{2} \right)-2{{\tan }^{-1}}x+C\]

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\[\int{\frac{{{x}^{2}}}{({{x}^{2}}+1)({{x}^{2}}+4)}dx}\] is equal to

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