\[\int_{{}}^{{}}{{{e}^{{{\tan }^{-1}}x}}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)\ dx\] is equal to

\[{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}+c\]

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

\[\int_{{}}^{{}}{{{e}^{{{\tan }^{-1}}x}}}\left( \frac{1+x+{{x}^{2}}}{1

1.	\[\int_{{}}^{{}}{{{e}^{{{\tan }^{-1}}x}}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)\ dx\] is equal to
A.	\[x{{e}^{{{\tan }^{-1}}x}}+c\]
B.	\[{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}+c\]
C.	\[\frac{1}{x}{{e}^{{{\tan }^{-1}}x}}+c\]
D.	None of these
Answer» B. \[{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}+c\]

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\[\int_{{}}^{{}}{{{e}^{{{\tan }^{-1}}x}}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)\ dx\] is equal to

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