Making a detour

Yes, exactly (pun intended)! (+1)

For the members who are not so familiar with the subject, we should point out that a potential of the given 1-form $\omega=\frac{x}{(x^2+y^2)^{3/2}}dx+\frac{y}{(x^2+y^2)^{3/2}}dy$ is $f(x,y)=-\frac{1}{\sqrt{x^2+y^2}}$ , meaning that $df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\omega$ . As you say, the integral depends only on the initial point $A=(1,0)$ and the terminal point $B=(2,0)$ of the path $C$ ; it is "path-independent". Then $\int_C df=f(B)-f(A)=\frac{1}{2}$ .

For those who are familiar with single-variable calculus only: The potential $f$ plays the role of an antiderivative, and $\int_C df=f(B)-f(A)$ is a version of the fundamental theorem, analogous to the single-variable case.

In terms of physics, we are looking at an inverse square field here that could represent a gravitational field, while $f(x,y)$ could represent the gravitational potential energy (up to a constant multiple, of course).