In an engineering textbook I found this formula for "the potential $u$ of a vector field $V=(X,Y,Z)$.
$u=u(x,y,z) = \int_{x_1}^x X(x,y,z) dx + \int_{y_1}^y X(x,y,z) dy + \int_{z_1}^z X(x,y,z) dz$
My Guess is from point $(x_1,y_1,z_1)$ to any point $(x,y,z)$ if $V$ is a gradient field. But even in this case the formula is not correct or is it?
The formula is correct if the region of integration is simply connected and $\nabla\times V=0$ on this region. See this article in Wikipedia. Another way to state the same result is that a vector field has a single-valued potential over some region if the line integrals over any closed curve is zero. The classical example of a vector field (in fact, a differential form) that has zero curl over a region, but does not have a single-valued potential is given in this question.