\[\int_{{}}^{{}}{\frac{{{e}^{m{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}\] equals to [RPET 2001]

\[\frac{1}{m}{{e}^{{{\tan }^{-1}}x}}\]

\[\frac{1}{m}{{e}^{m{{\tan }^{-1}}x}}\]

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

1.	\[\int_{{}}^{{}}{\frac{{{e}^{m{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}\] equals to [RPET 2001]
A.	\[{{e}^{{{\tan }^{-1}}x}}\]
B.	\[\frac{1}{m}{{e}^{{{\tan }^{-1}}x}}\]
C.	\[\frac{1}{m}{{e}^{m{{\tan }^{-1}}x}}\]
D.	None of these
Answer» D. None of these

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\[\int_{{}}^{{}}{\frac{{{e}^{m{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}\] equals to [RPET 2001]

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