A) \[y=\log (1+{{x}^{2}})+c\]
B) \[y+\log (1+{{x}^{2}})+c=0\]
C) \[y-\log (1+x)=c\]
D) \[y={{\tan }^{-1}}x+c\]
Correct Answer: D
Solution :
\[(1+{{x}^{2}})\frac{dy}{dx}=1\]Þ\[\frac{dy}{dx}=\frac{1}{1+{{x}^{2}}}\] On integrating, \[y={{\tan }^{-1}}x+c\].You need to login to perform this action.
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