= Solution
For a planar equilibrium, multiply $\mu=0$ by $\phi'$ and use the <chain rule>. The result is
$$
\frac d{dx}\left[
f(\phi)-\frac{\kappa(\phi)}2(\phi')^2
\right]=0.
$$
Matching to either bulk phase fixes the constant:
$$
\boxed{
f(\phi)-\frac{\kappa(\phi)}2(\phi')^2=f(\phi_B)
}.
$$
Consequently
$$
\frac{dx}{d\phi}
=\pm\sqrt{\frac{\kappa(\phi)}
{2[f(\phi)-f(\phi_B)]}},
$$
and the inverse profile is
$$
\boxed{
x(\phi)-x_0
=\pm\int_0^\phi
\sqrt{\frac{\kappa(\psi)}
{2[f(\psi)-f(\phi_B)]}}\,d\psi
}.
$$
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