How is this Taylor expansion computed? I am reading a paper, where we consider the following function,
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)$$
where $b>0$ is a constant such that $r>2/b.$ After stating this definition the author writes that the taylor expression of $F$ is,
$$F(r) = b\frac{r^2}{4}-\frac{\ln(r)}{b} -\frac{1}{2b} -\frac{\ln(2b)}{b} + \mathcal{O}(b^{-3}r^{-2}).$$
I know how to compute taylor series, but I am not sure what the taylor series is centered around. I have computed the derivatives upto order $2$. 
$$F'(r) = \frac{1}{2}\sqrt{b^2r^2-4}, \quad F''(r) = \frac{b^2r}{2\sqrt{b^2r^2-4}}.$$ 
Any explanations regarding the expansion would be much appreciated. 

Here you need the following expansions centered at $x=0$:
$$\sqrt{1+x}=1+\frac{x}{2}+O(x^2)\quad,\quad\ln(1+x)=x+O(x^2).$$
Hence, for $r$ large enough and $x=-\frac{4}{b^2r^2}$,
$$\begin{align*}F(r)&=\frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{1}{b}\ln(br(\sqrt{1-\frac{4}{b^2r^2}} + 1))\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{\ln(br)}{b}-\frac{1}{b}\ln(\sqrt{1-\frac{4}{b^2r^2}}+1)\\
&=\frac{br^2}{4}\left(1-\frac{4}{2b^2r^2}\right)+O(b^{-3}r^{-2})-\frac{\ln(br)}{b}-\frac{1}{b}\ln(2-\frac{4}{2b^2r^2}+O(b^{-4}r^{-4}))\\
&=\frac{br^2}{4}-\frac{1}{2b}+O(b^{-3}r^{-2})-\frac{\ln(br)}{b}-\frac{\ln(2)}{b}-\frac{1}{b}\ln(1-\frac{1}{b^2r^2}+O(b^{-4}r^{-4}))\\
&=\frac{br^2}{4}-\frac{1}{2b}-\frac{\ln(2b)}{b}+\frac{\ln(r)}{b}+O(b^{-3}r^{-2})
\end{align*}$$
