Integrating the top Chern class, where am I going wrong?

Consider the vector bundle $$\mathcal{E}\rightarrow\mathcal{D}$$ over the open unit complex disc, where the fiber over $$z\in\mathcal{D}$$ is the span of the vector $$(z, 1)$$. Consider the Hermitian metric given by $$\langle(z, 1), (z, 1)\rangle=1-|z|^2$$.
Thus the connection associated to the metric is $$\omega=\frac{\bar{z}dz}{|z|^2-1}$$, and the curvature is $$\Omega=\frac{dz\wedge d\bar{z}}{(|z|^2-1)^2}$$.
The top Chern class is represented by $$c_1=\frac{-dz\wedge d\bar{z}}{2\pi i(|z|^2-1)^2}$$.
Now integrating this over $$\mathcal{D}$$ we first use $$dz =dx+idy$$ and $$d\bar{z}=dx-idy$$ to switch to real coordinates, and after a change to polar coordinates we should have the integral $$\int_0^1\int_0^{2\pi}\frac{r drd\theta}{\pi(r^2-1)^2}$$. Integrating over $$\theta$$ and then setting $$r^2-1=u$$ we have $$\int_{-1}^0\frac{du}{u^2}$$ which is divergent.
My issues is that since $$\mathcal{D}$$ is contractible, every bundle should be trivial. Unless I'm mistaken, if the integral of the top Chern class is non-zero, then the bundle can not be trivial.
I'm also fairly sure that I read that the integral of the top Chern class of a line bundle should be an integer. So considering this, what have I done wrong?

• The theory you're relying on is for compact manifolds, no? – Ted Shifrin May 28 at 18:19
• I'm not sure, but that would make sense, as all the examples I've seen of this in use are in fact compact. – Kristaps John Balodis May 28 at 19:23