The solution of the differential equation\[\frac{dy}{dx}+\frac{2yx}{1+

1.	The solution of the differential equation\[\frac{dy}{dx}+\frac{2yx}{1+{{x}^{2}}}=\frac{1}{{{(1+{{x}^{2}})}^{2}}}\] is:
A.	\[y(1+{{x}^{2}})=c+{{\tan }^{-1}}x\]
B.	\[\frac{y}{1+{{x}^{2}}}=c+{{\tan }^{-1}}x\]
C.	\[y\log (1+{{x}^{2}})=c+{{\tan }^{-1}}x\]
D.	\[y(1+{{x}^{2}})=c+{{\sin }^{-1}}x\]
Answer» B. \[\frac{y}{1+{{x}^{2}}}=c+{{\tan }^{-1}}x\]

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The solution of the differential equation\[\frac{dy}{dx}+\frac{2yx}{1+{{x}^{2}}}=\frac{1}{{{(1+{{x}^{2}})}^{2}}}\] is:

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