How can I solve $(xy^3 + y)dx + 2(x^2y^2 + x + y^4)dy = 0$? 
Solve this differential equation
  $$(xy^3 + y)dx + 2(x^2y^2 + x + y^4)dy = 0$$

I tried converting it to the form: $\frac{dy}{dx} + yp(x) = q(x)$ but couldn't. The equation is also not homogeneous. Keeping $\frac{dy}{dx}$ on one side will not render the numerator as the derivative of the denominator (with some manipulation) on the other side of the equation.

By multiplying both sides by $y$ (see Moo's comment) we have that
$$0=y(xy^3 + y)dx + 2y(x^2y^2 + x + y^4)dy=d\left(\frac{y^2(2y^4+3x^2y^2+6x)}{6}\right).$$
Hence
$$y^2(2y^4+3x^2y^2+6x)=C$$
where $C$ is an arbitrary constant. Have you any initial condition?
