In this post: How to integrate a total derivative?,
Dylan says that to integrate
$$df = x \hphantom{.} dy + y \hphantom{.} dx $$
You can use the following notation:
$$ \int \frac{\partial f}{\partial y}\ dy = \int x \ dy + \int y\frac{\partial x}{\partial y} dy $$
$$ f(x, y) = [xy + g(x)] + [0]$$
When integrating a differential form with $dx$ and $dy$, you must first work with either $\frac{\partial F}{\partial y}$ or $\frac{\partial F}{\partial x}$ in isolation, and say you integrate $\frac{\partial F}{\partial y}$ first, then you must work on $\frac{\partial F}{\partial x}$ through $g(x)$ in the expression $f(x, y) = \int \frac{\partial F}{\partial y}dy + g(x)$.
So Dylan seems to be presenting a notational trick to act on the differential form in such a way to get $\frac{\partial F}{\partial y}$ alone by wiping out $\frac{\partial F}{\partial x}$. The purpose of replacing $ \hphantom{.}y \hphantom{.} dx \hphantom{.}$ with $ \hphantom{.} y\frac{\partial x}{\partial y} dy \hphantom{.}$ is to make it go to zero, so that you can work on $\frac{\partial F}{\partial y}$ in isolation. It's not needed, but I guess it's a notational preference if on paper you want integrate from the original equation.
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The RHS makes sense to me, but in the LHS, I don't the substation he makes: $df = \frac{\partial f}{\partial y}dy$, since df only partially depends on y. To me, it seems rather that $\frac{\partial f}{\partial y}\ dy = df - \frac{\partial f}{\partial x}\ dx$.
Is what he is saying correct? Or am I misunderstanding something?
And if it's not correct, it's too bad that integration of a 1-form with multiple variables requires you to start a new section on a piece of paper, there isn't any flow where you place an equation in $\int ( \cdot ) $ and line by line work out the answer. Instead you are moving from $\int \frac{\partial f}{\partial z} dz + H(x, y)$ to $\int \frac{\partial f}{\partial y} dz + G(x)$ to $\int \frac{\partial f}{\partial x} dz + C$
Thanks.
Basically it's a matter of notation. Some people use $\dfrac{\partial f}{\partial y}$ to indicate the "total derivative with respect to $y$". What they really mean is that if $f$ is a function of $x$ and $y$, then you can parameterize one of them with respect to the other. Then $f$ composed with that parameterization, for example, $f(x(y),y)$, is a function of $y$ only. As such, the differential is $$df=f'(y)dy \hspace{2mm}\text{or}\hspace{2mm}\frac{df}{dy}dy \hspace{2mm} \text{or} \hspace{2mm}\frac{\partial f}{\partial y}dy$$ I prefer the first two notations because they are inequivocal as to what kind of derivative is being taken, and what paremeters exactly is $f$ a function of.
You can also note that using that preferred notation one can also use chain rule to write $$df=\left(\frac{\partial f}{\partial x}x'(y)+\frac{\partial f}{\partial y}\right)dy$$ and noting that $x'(y)dy=dx$ (this is also what they use for the rightside integral) for that parametrization you again arrive at $df=\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial x}dx$.
So the moral of the story is that many people have different notations for the same thing. Whenever you see this $\frac{\partial f}{\partial x}$ beware that it can mean different things.
EDIT:
As for your integral, like one of the answers stressed, forms are not only "integrated" in an abstract; they are integrated over a path. So you can in fact integrate the 1-form "in one go", you just have to choose a suitable path, like he did. Dylan's method also works but he technically wasn't integrating a differential form: he knew it was exact, and from that knowledge he set up a system of differential equations and solved for $f$, which is why he integrates one thing at a time.