Solving easy first-order linear differential question. Question
Solve $y'=2x(1+x^2-y)$.
My attempt
Rearranging gives $y'+2xy=2x(1+x^2)$. Thus, the integrating factor is $e^{\int2x\,dx}=e^{x^2}$ and multiplying the equation throughout by this gives
$e^{x^2}y'+2xe^{x^2}y=2xe^{x^2}(1+x^2)\Rightarrow\dfrac{d}{dx}{e^{x^2}y}=2xe^{x^2}+2x^3e^{x^2}$ which is separable.
Then,
$e^{x^2}y=\int2xe^{x^2}\,dx+2\int x^3e^{x^2}\,dx=e^{x^2}+2\int x^3e^{x^2}\,dx$
Is this correct and is there some easy way of working out $\int x^3e^{x^2}\,dx$? Wolfram has it as $e^{x^2}(x^2-1)$.
Maybe I write $2x^3e^{x^2}$ as $x(2x^2e^{x^2})$?

make a change of variable $$1+x^2 - y = u, \quad y =1+x^2 - u, y' = 2x-u' $$ then the de $y' = 2x(1+x^2 - y)$ is turned into $$2x-u' = 2xu $$ multiplying by $e^{x^2},$  we get $$e^{x^2}(u'+2xu) = \left(e^{x^2}u\right)' = 2xe^{x^2}$$ on integration yields $$e^{x^2}u = e^{x^2} + c\to u = 1+ce^{-x^2},\quad y = x^2 -ce^{-x^2} $$
