 Solution

Hydrostatic pressure varies with height as $p_l-p_{\mathrm{atm}}=-\rho gz$ when the far flat bath is the zero-height reference. Surface tension balances that pressure jump by curvature; with the curvature sign chosen in the equation this becomes $\rho gz=\sigma \kappa$. Here $\rho$ is liquid mass density, $g$ gravitational acceleration, $\sigma$ surface tension and $\kappa$ the signed planar curvature. The capillary length ${\ell }_c=\sqrt{\sigma /(\rho g)}$ balances these effects. Scale both physical $z$ and $h$ by ${\ell }_c$ to obtainFor the wall-rising meniscus $h$ decreases as $z$ rises. At the distant bath, $z\downarrow 0$, $h\to \infty$ and $h_z\to -\infty$, because the free surface becomes horizontal when expressed as $z(x)$. In the first integral this gives $\boxed{A=-1}$.

At zero wall contact angle the interface is tangent to the vertical wall. In the representation $x=h(z)$ this means $h(z_0)=0$ and $h_z(z_0)=0$, so $z_0^2/2-1=0$. The rising contact point is $\boxed{z_0=\sqrt{2}}$. Put $q=z^2/2-1$. Then $h_z=q/\sqrt{1-q^2}=\sqrt{2}(z-z_0)+O((z-z_0{)}^2)$. Integration from the contact point givesThe sign of $h_z$ is negative on the approach from below, while $h$ is positive, as required by the geometry. This local quadratic shape is the zero-contact-angle wall meniscus matching law.