A) \[2{{\tan }^{-1}}y=\log (1+{{x}^{2}})+c\]
B) \[{{\tan }^{-1}}y=\log (1+{{x}^{2}})+c\]
C) \[2{{\tan }^{-1}}y+\log (1+{{x}^{2}})+c=0\]
D) None of these
Correct Answer: A
Solution :
\[(1+{{x}^{2}})\frac{dy}{dx}=x(1+{{y}^{2}})\]Þ\[\frac{1}{1+{{y}^{2}}}dy=\frac{x}{1+{{x}^{2}}}dx\] On integrating, we get \[{{\tan }^{-1}}y=\frac{1}{2}\log (1+{{x}^{2}})+c\] Þ \[2{{\tan }^{-1}}y=\log (1+{{x}^{2}})+c\].
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