differential linear equation of order one $(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$ I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$
Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$
It is not exact since partial derivatives are not equal.
Any help would be appreciated.

To put it into the form you requested:
$$
-(1+x^2) \,dy + (2xy + x^2 + x^4) \,dx = 0 \implies \frac{dy}{dx} - \frac{2xy + x^2 + x^4}{1+x^2} = 0 \\
\implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = \frac{x^2 + x^4}{1+x^2}\\
\implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = x^2
$$
