= Solution
The <gradient> part contributes zero around a closed path by the <fundamental theorem for line integrals>:
$$
\oint_{\Gamma_1}3\nabla\phi\cdot d\mathbf x=0.
$$
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 $\mathbf w\cdot d\mathbf x=x\,dz=\cos^2\theta\,d\theta$. Thus
$$
\boxed{\oint_{\Gamma_1}(3\mathbf v+\mathbf w)\cdot d\mathbf x
=\int_0^{2\pi}\cos^2\theta\,d\theta=\pi}.
$$
\b[The printed field needs a qualification:] $\operatorname{arctanh}(x^2)$ is undefined at the isolated parameters where $x^2=1$, so $0\operatorname{arctanh}(1)$ 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.
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