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) \[{{e}^{x}}\left[ \log (1+{{x}^{2}})-\frac{x}{1+{{x}^{2}}} \right]\]
Correct Answer: A
Solution :
\[\frac{d}{dx}\{{{e}^{x}}\log (1+{{x}^{2}})\}={{e}^{x}}\log (1+{{x}^{2}})+{{e}^{x}}\frac{1}{(1+{{x}^{2}})}2x\] \[={{e}^{x}}\left[ \log (1+{{x}^{2}})+\frac{2x}{1+{{x}^{2}}} \right]\].
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