Line integral of a vector potential From the theory of electromagnetism, the line integral $\int {\bf A}\cdot{d{\bf s}}$ is independent of paths, that is, it is dependent only on the endpoints, as long as the loop formed by pair of different paths does not enclose a magnetic flux.

Why is this true?
 A: From a slightly different, though equivalent, view...
If the line integral of a vector field is path independent, the vector field is conservative, i.e., the vector field is the gradient of a scalar field.
Thus, if $\int \mathbf{A}\cdot \mathrm{d}\mathbf{s}$ is path independent, it is the case that $\mathbf{A} = \nabla \phi$.
Now, recall that the curl of a divergence is identically zero:
$$\nabla \times \nabla\phi = \mathbf{0}$$
But, the magnetic field is $\mathbf{B} = \nabla \times \mathbf{A}$ and thus, in this case, $\mathbf{B} =0$
A: It's a straightforward application of Stokes' theorem:
Given two paths $\gamma_1,\gamma_2$ with the same starting and end points, let $\gamma := \gamma_1 - \gamma_2$ be the loop obtained by going from the starting point along $\gamma_1$ to the end point, and then in the reverse direction along $\gamma_2$. Let $S$ be a surface filling $\gamma$, i.e. such that its boundary is $\gamma$.
Then we have that
$$ \int_{\gamma_1} A - \int_{\gamma_2} A = \int_\gamma A = \int_{\partial S} A = \int_S \mathrm{d}A$$
and in vector notation $\mathrm{d}A$ is $\nabla\times A = B$. But $\int_S B$ is just the magnetic flux through $S$, so if there's no flux enclosed by $\gamma$, the two integrals along the paths are equal.
A: No magnetic flux means :
$$\int \vec B \cdot d \vec a=0$$ 
implies 
$$\int (\nabla\times \vec A )\cdot d \vec a = 0$$
By Stoke theorem:
$$\oint \vec{A} \cdot{d\vec{l}}= 0$$
Which means $\vec A$ is path independent, as the area* is arbitrarily chosen, and a arbitrary loop integral is just sum of two arbitrary line integral (with overlapped starting and ending points).

*The integrals are over arbitrary area at where no flux, and the loop integral is over the boundary of the chosen area. 
