Let $\overline{M}$ be a projective manifold, $D$ is smooth divisor in $\overline{M}$. Suppose $S$ is the defining section of $D$, that is $S$ is a nontrivial meromorphic section of the line bundle $[D]$ such that $D=\{S=0\}$. If we further suppose that $[D]$ is an ample line bundle, then it's direct to compute that $\omega_g=\frac{\sqrt{-1}}{2\pi}\partial\overline{\partial}(-\log\vert|S\vert|^2)^{\frac{n+1}{n}}$ is a Kahler metric on $M=\overline{M}\backslash D$. The problem is that if we fix a point $x_0\in M$, then when $x$ is sufficiently close to $D$, the distance function $r(x)=d(x,x_0)$ is of order $O((-\log\vert|S\vert|^2)^{\frac{n+1}{2n}})$, where $\vert|\cdot\vert|$ is some Hermitian metric on line bundle $[D]$.
This claim comes from the paper "Complete Kahler Manifolds with Zero Ricci Curvature", and I'm confused with this. Although the setting is complex, the problem is essentially Riemannian geometry. As far as I'm concerned, a standard process of caculating the distance function involves finding a geodesic $\gamma(t)$ connecting $x$ and $x_0$, then use $d(x,x_0)=\int_0^1\vert|\gamma'(t)\vert|dt$. However, in our setting, it's too complicated to carry out this process since finding a geodesic explicitly is already horrifying. Still the claim does not require a precise formula of distance function, so this may rather be a simple question where I got stuck at some stupid point. Any solution or hint is highly appreciated!