Neglected terms in integral sum As is known, we can neglect high-order term in expression $f(x+dx)-f(x)$. For example, $y=x^2$: $dy=2xdx+dx^2$, $dy=2xdx$.
I read that infinitesimals have property: $dx+dx^2=dx$
I tried to neglect high-order terms in integral sum ($dx^2$ and $4dx^2$ and so on) and I obtained wrong result in the end.

Where is my mistake?
Thanks
 A: There are many different answers depending on the way in which one makes arguments about infinitesimals are rigorous. Since your question did not pick a particular way, my answer won't either.

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*"$\mathrm dx+\mathrm dx^2=\mathrm dx$" is almost never something we would consider true. What is often true is that if you have something like $\dfrac{3\mathrm dx+\mathrm dx^2}{\mathrm dx}$, then you probably want to simplify that down to $3+\mathrm dx$, and then only care about the $3$.

*When you have something like the sum of terms like $(x_0^2+6x_0\,\mathrm dx+9\mathrm dx^2)\mathrm dx$ (which you'd probably want to simplify to $x_0^2\mathrm dx+6x_0\,\mathrm dx^2+9\mathrm dx^3$), you need to keep in mind that if there are $N$ terms like that, then the sum of the $\mathrm dx^3$ subterms has a coefficient on the order of $N^3$ because you are summing the first $N$ squares. Thus, that part gives you something in the ballpark of $N^3\mathrm dx^3$, which might be something like $(x_a-x_0)^3$, which can't be discarded since there's no factor $\mathrm dx$ left over.

