Prove that the line integral of a vector valued function does not depend on the particular path Let C denote the path from $\alpha$ to $\beta$.
  If $\textbf{F}$ is a gradient vector, that is, there exists a differentiable function $f$ such that
$$\nabla f=F,$$ then
\begin{eqnarray*}
\int_{C}\textbf{F}\; ds &=& \int_{\alpha}^{\beta} \textbf{F}(\vec{c}(t)).\vec{c'}(t)\; dt \\
&=&\int_{\alpha}^{\beta} \nabla f(\vec{c}(t)).\vec{c'}(t)\; dt \\
&=&\int_{\alpha}^{\beta}  \frac{\partial  f(\vec{c}(t))}{\partial t}\; dt \\
&=&f(\vec{c}(\beta))- f(\vec{c}( \alpha))
\end{eqnarray*}
That is, the integral of $\textbf{F}$ over $C$ depends on the values of the end points $c(\beta)$ and $c(\alpha)$ and is thus independent of the path between them.
This proof is true if and only if  $\textbf{F}$ is a gradient vector, what if not ?
 A: If F is not a gradient vector field, there is no potential function to evaluate at the end points and the penultimate and last steps fail. The proof that path is irrelevant when there is a potential function relies on the fundamental theorem for line integrals, which no longer applies if there is no primitive. 
For the proof, refer to my answer here:
Is path length somehow irrelevent in this question? Line integrals Calc 3
A: In general, for a smooth vector field $\vec F(\vec r)$, Helmholtz's Theorem guarantess that there exists a scalar field $\Phi(\vec r)$ and a vector field $\vec A(\vec r)$ such that
$$\vec F(\vec r)=\nabla \Phi(\vec r)+\nabla \times \vec A(\vec r)$$
Then, forming the line integral along a path $C$, from $\vec r_1$ to $\vec r_2$, we see that
$$\begin{align}
\int_C \vec F(\vec r)\cdot \,d\vec \ell&=\int_C\left(\nabla \Phi(\vec r)+\nabla \times \vec A(\vec r)\right)\cdot \,d\vec \ell\\\\
&=\Phi(\vec r_2)-\Phi(\vec r_1)+\int_C \nabla \times \vec A(\vec r) \cdot \,d\vec \ell \tag 1\\\\
\end{align}$$
Inasmuch as the integral on the right-hand side of $(1)$ is, in general, path dependent, then the path integral of $\vec F(\vec r)$ is also, in general, path dependent.
