\[\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}dx=}\] [MP PET 1991]

\[x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c\]

\[2x{{\tan }^{-1}}x+\log (1+{{x}^{2}})+c\]

\[2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c\]

\[\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}dx=}\]

1.	\[\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}dx=}\] [MP PET 1991]
A.	\[x{{\tan }^{-1}}x+c\]
B.	\[x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c\]
C.	\[2x{{\tan }^{-1}}x+\log (1+{{x}^{2}})+c\]
D.	\[2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c\]
Answer» E.

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\[\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}dx=}\] [MP PET 1991]

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