 Solution

The gradient part contributes zero around a closed path by the fundamental theorem for line integrals:Along the parametrization, $x^2+z^2=1$, so the first two components of $\mathbf{w}$ vanish wherever the printed expression is defined. The remaining component gives $\mathbf{w}\cdot d\mathbf{x}=xdz={\cos }^2\theta d\theta$. ThusThe printed field needs a qualification: $\mathrm{arctanh}(x^2)$ is undefined at the isolated parameters where $x^2=1$, so $0\mathrm{arctanh}(1)$ cannot literally be evaluated. The result above is the improper line integral, or equivalently the integral of the continuous extension of the restricted field along the curve. That restricted field tends to $(0,0,\cos \theta )$ at each omitted parameter. No extension of the full field to a neighbourhood is being assumed.