Is my solution to this differential equation correct? 
My answer is: $[(1+x^2)^3]y = \dfrac{(1+x^2)^3}3+C$
But this option is not given, so is it correct? 
Thanks

$$(1+x^2)\frac{dy}{dx}+6xy=2x \Rightarrow (1+x^2)\frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2)\frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{1}{1-3y}dy=\frac{2x}{1+x^2}dx \Rightarrow \int \frac{1}{1-3y}dy=\int \frac{2x}{1+x^2}dx \\ \Rightarrow \frac{-1}{3}\ln{|1-3y|}=\ln{|1+x^2|}+c\Rightarrow \ln{|1-3y|}^{\frac{-1}{3}}=\ln{|1+x^2|}+c \\ \Rightarrow e^{\ln{|1-3y|}^{\frac{-1}{3}}}=e^{\ln{|1+x^2|}+c } \\ \Rightarrow |1-3y|^{\frac{-1}{3}}=C|1+x^2|, C=e^c$$
Solve now for $y$.
