I'm currently trying to find the variance stabilizing transformation of a random variable $N\sim\text{NB}(1,p)$, i.e. $\mathbb{P}(N=n)=p(1-p)^n$ for $n=0,1,2,...$, and I'm a bit stuck.
Through $\ell (p\, ;\, n) = \log p - n\log (1-p)$ I've found the Fisher information to be
$$I_p(p)=\mathbb{E}_p[I(p\,;\,N)] = \mathbb{E}_p\bigg[-\frac{\partial^2}{\partial p^2}\ell(p\,;\,N)\bigg]=\frac{1}{p^2}-\frac{p}{(1-p)^3}.$$
Using the formula from my book, I'm supposed to get the variance stabilizing transformation using
$$\eta =\eta(p)= \frac{1}{c} \int_{\cdot}^p \sqrt{I_p(y)}\,dy.$$
This transformation is going to give me a constant information $c^2$.
From this paper (Proposition 1) I can sense I'm suppose to use some smart substitution trick in order to get the inverse hyperbolic sine function. Can someone point me in some direction in order to calculate this and get a rather nice expression like in the paper?
Edit: I found out that I was using the wrong parametrization. We recognize $\mathbb{P}(N=n)=p(1-p)^n$ as a geometric distribution and hence $\mathbb{E}[N] = \frac{1-p}{p}$ and $Var(N) = \frac{1-p}{p^2}$. Thus the Fisher information is
$$I_p(p) = \frac{1}{p^2(1-p)}.$$
I've added my own answer down below.
Note the Edit in the question above.
Using the formula together with this Fisher information, we obtain
$$\begin{align*}\eta &= \frac{1}{c} \int_{\cdot}^p \sqrt{\frac{1}{y^2(1-y)}}\ dy \\ &= \frac{1}{c} \int_{\cdot}^p \frac{1}{y}\frac{1}{\sqrt{1-y}}\ dy \end{align*}$$
Through integration by substitution by letting $u=\sqrt{1-y}$ so $y=1-u^2$, we get that this intergral is
$$ \frac{1}{c} \int_{\cdot}^p \frac{1}{y}\frac{1}{\sqrt{1-y}}\ dy = \frac{1}{c}\Big[2\text{tanh}^{-1}(\sqrt{1-y})\Big]_{\cdot}^p = \frac{2}{c}\,\text{tanh}^{-1}(\sqrt{1-p})$$
So, for $\eta = \frac{2}{c}\text{tanh}^{-1}(\sqrt{1-p})$ we obtain constant information $c^2$.