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
 A: $$(1+x^2) \frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2) \frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{dy}{1-3y}=\frac{2x}{1+x^2} dx \\ \Rightarrow -\frac{1}{3} \ln |1-3y|=\ln |1+x^2|+c$$
A: $$(1+x^2)\frac{dy}{dx}+6xy=2x \Rightarrow (1+x^2)\frac{dy}{dx}=2x-6xy \Rightarrow (1+x^2)\frac{dy}{dx}=2x(1-3y) \Rightarrow \frac{1}{1-3y}dy=\frac{2x}{1+x^2}dx \Rightarrow \int \frac{1}{1-3y}dy=\int \frac{2x}{1+x^2}dx \\ \Rightarrow \frac{-1}{3}\ln{|1-3y|}=\ln{|1+x^2|}+c\Rightarrow \ln{|1-3y|}^{\frac{-1}{3}}=\ln{|1+x^2|}+c \\ \Rightarrow e^{\ln{|1-3y|}^{\frac{-1}{3}}}=e^{\ln{|1+x^2|}+c } \\ \Rightarrow |1-3y|^{\frac{-1}{3}}=C|1+x^2|, C=e^c$$
Solve now for $y$.
A: $$\begin{align}
(1+x^2)\frac{dy}{dx}+6xy&=2x\\
(1+x^2)\frac{dy}{dx}&=2x(1-3y)\\
\int \frac{dy}{1-3y}&=\int\frac {2x}{1+x^2}dx\\
-\frac13\ln(1-3y)&=\ln(1+x^2)\\
(1-3y)(1+x^2)^3&=0\\
(1+x^2)^3&=3y(1+x^2)^3\\
1+3x^2+3x^4+x^6&=3y(1+x^2)^3\\
\frac 13+x^2+x^4+\frac{x^6}3&=y(1+x^2)^3\\
y(1+x^2)^3&=x^2+x^4+\frac{x^6}3+C
\end{align}$$
i.e. option (A) $\blacksquare$.
Check by differentiating:
$$\begin{align}
(1+x^2)^3\frac{dy}{dx}+6xy(1+2x^2)^2&=2x+4x^3+2x^5\\
&=2x(1+2x^2+x^4)\\
&=2x(1+x^2)^2\\
(1+x^2)\frac{dy}{dx}+6xy&=2x
\end{align}$$
which is the original equation, hence solution is correct. 
