\[\int_{{}}^{{}}{\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=}\] [MP PET 1987]

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

\[\log {{e}^{{{\tan }^{-1}}x}}+c\]

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

1.	\[\int_{{}}^{{}}{\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=}\] [MP PET 1987]
A.	\[\log (1+{{x}^{2}})+c\]
B.	\[\log {{e}^{{{\tan }^{-1}}x}}+c\]
C.	\[{{e}^{{{\tan }^{-1}}x}}+c\]
D.	\[{{\tan }^{-1}}{{e}^{{{\tan }^{-1}}x}}+c\]
Answer» D. \[{{\tan }^{-1}}{{e}^{{{\tan }^{-1}}x}}+c\]

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\[\int_{{}}^{{}}{\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=}\] [MP PET 1987]

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