\[\frac{d}{dx}[(1+{{x}^{2}}){{\tan }^{-1}}x]=\] – McqOptions

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1.	\[\frac{d}{dx}[(1+{{x}^{2}}){{\tan }^{-1}}x]=\]
A.	\[x{{\tan }^{-1}}x\]
B.	\[2{{\tan }^{-1}}x\]
C.	\[2x{{\tan }^{-1}}x+1\]
D.	\[x{{\tan }^{-1}}x+1\]
Answer» D. \[x{{\tan }^{-1}}x+1\]

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