\[\frac{d}{dx}\left\{ {{e}^{x}}\log (1+{{x}^{2}}) \right\}=\] [AI CBSE 1987]

\[{{e}^{x}}\left[ \log (1+{{x}^{2}})-\frac{2x}{1+{{x}^{2}}} \right]\]

\[{{e}^{x}}\left[ \log (1+{{x}^{2}})+\frac{2x}{1+{{x}^{2}}} \right]\]

\[{{e}^{x}}\left[ \log (1+{{x}^{2}})+\frac{x}{1+{{x}^{2}}} \right]\]

\[{{e}^{x}}\left[ \log (1+{{x}^{2}})-\frac{x}{1+{{x}^{2}}} \right]\]

\[\frac{d}{dx}\left\{ {{e}^{x}}\log (1+{{x}^{2}}) \right\}=\] �

1.	\[\frac{d}{dx}\left\{ {{e}^{x}}\log (1+{{x}^{2}}) \right\}=\] [AI CBSE 1987]
A.	\[{{e}^{x}}\left[ \log (1+{{x}^{2}})+\frac{2x}{1+{{x}^{2}}} \right]\]
B.	\[{{e}^{x}}\left[ \log (1+{{x}^{2}})-\frac{2x}{1+{{x}^{2}}} \right]\]
C.	\[{{e}^{x}}\left[ \log (1+{{x}^{2}})+\frac{x}{1+{{x}^{2}}} \right]\]
D.	\[{{e}^{x}}\left[ \log (1+{{x}^{2}})-\frac{x}{1+{{x}^{2}}} \right]\]
Answer» B. \[{{e}^{x}}\left[ \log (1+{{x}^{2}})-\frac{2x}{1+{{x}^{2}}} \right]\]

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\[\frac{d}{dx}\left\{ {{e}^{x}}\log (1+{{x}^{2}}) \right\}=\] [AI CBSE 1987]

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