Htykuut

I'm a third year physics student, not a math student.
The conventional way to solve this would be:
$$frac{d^2u}{dx^2}=frac{d}{dx}(frac{du}{dx})=1 Rightarrow d(frac{du}{dx})= dxRightarrow int d(frac{du}{dx})=int dxRightarrow frac{du}{dx}=x+A Rightarrow
du=(x+A)dx Rightarrow u=frac{x^2}{2}+Ax+B$$

So I'm wondering if what follows is correct:
$$frac{d^2u}{dx^2}=1 Rightarrow d^2u=(dx)^2 Rightarrow d(du)=dxdx Rightarrow int d(du)=int dxdx Rightarrow du=int dxdx$$

At this point, I don't know if there are other ways to proceed, but I use integration by parts
$$int vdw=vw-int wdv$$
by considering $v=dx$ and $dw=dx$, so that $w=x$ and $dv=d(dx)$:
$$du=int dxdx=xdx-int x·d(dx)$$

If I assume that $-int x·d(dx)=C$ is a constant, we get
$$du=xdx+C Rightarrow u=frac{x^2}{2}+Cx+D$$
which is what we had originally.
However, how would I go about proving that $int x·d(dx)=constant$? Is it because $d(dx)=0$? And how do I prove that? Are there any other interesting ways to solve this differential equation?

Sorry, this is not correct. Already the first step $d^2 u = (dx)^2$ is meaningless.