A) \[y={{\tan }^{-1}}x+c\]
B) \[y=-{{\tan }^{-1}}x+c\]
C) \[y=\frac{1}{2}{{\log }_{e}}(1+{{x}^{2}})+c\]
D) \[y=-\frac{1}{2}{{\log }_{e}}(1+{{x}^{2}})+c\]
Correct Answer: C
Solution :
\[(1+{{x}^{2}})\frac{dy}{dx}=x\]Þ \[dy=\frac{x}{1+{{x}^{2}}}dx\] Þ \[\int{dy}=\int{\frac{x}{1+{{x}^{2}}}dx}+c\] Þ \[y=\frac{1}{2}{{\log }_{e}}(1+{{x}^{2}})+c\].
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