The value of \[\int{{{e}^{ta{{n}^{-1}}}}^{x}\frac{(1+x+{{x}^{2}})}{1+{{x}^{2}}}dx}\] is

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

The value of \[\int{{{e}^{ta{{n}^{-1}}}}^{x}\frac{(1+x+{{x}^{2}})}{1+{

1.	The value of \[\int{{{e}^{ta{{n}^{-1}}}}^{x}\frac{(1+x+{{x}^{2}})}{1+{{x}^{2}}}dx}\] is
A.	\[x{{e}^{{{\tan }^{-1}}}}x+c\]
B.	\[{{\tan }^{-1}}x+C\]
C.	\[{{e}^{{{\tan }^{-1}}x}}+2x+C\]
D.	None of these
Answer» B. \[{{\tan }^{-1}}x+C\]

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The value of \[\int{{{e}^{ta{{n}^{-1}}}}^{x}\frac{(1+x+{{x}^{2}})}{1+{{x}^{2}}}dx}\] is

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