Series expansion of $x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$ at $x=1$ In various texts it is stated that a good approximation of,
$x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$,
about $x=1$ is given by,
$\frac{4\sqrt{2}}{3} (x-1)^{3/2}$.
Plotting the graphs this is indeed the case, however, I am not sure how this approximation is derived. Entering the original equation in Wolfram Alpha it does list the above approximation as the first term in the Puiseux series expansion of the original function. I am not familiar with these and looking online I can't seem to find a way into calculating a Puiseux series for the relevant expression.
Any suggestions or pointers for how to go about this would be greatly appreciated :)
 A: The Approximation in the Question
I recognized this as the result of an integral that uses the trigonometric substitution $x=\sec(\theta)$, so I took the derivative of ${\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}$ and got $2{\textstyle\sqrt{t^2-1}}$. Thus,
$$
{\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}
=2\int_1^x{\textstyle\sqrt{t^2-1}}\,\mathrm{d}t\tag1
$$
For $t\approx1$, $t^2=(1+(t-1))^2\approx1+2(t-1)=2t-1$; therefore, $\sqrt{t^2-1}\approx\sqrt{2t-2}$. Thus,
$$
\begin{align}
\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)
&\approx2\int_1^x\sqrt{2t-2}\,\mathrm{d}t\tag2\\[6pt]
&=\textstyle\frac{4\sqrt2}3(x-1)^{3/2}\tag3
\end{align}
$$

A Better Approximation
We can get a better approximation if we note that
$$
\begin{align}
\textstyle2\sqrt{t^2-1}
&=\textstyle2\sqrt{2(t-1)+(t-1)^2}\tag{4a}\\
&=\textstyle2\sqrt2\sqrt{t-1}\sqrt{1+\frac{t-1}2}\tag{4b}\\
&\approx\textstyle2\sqrt2\sqrt{t-1}\left(1+\frac{t-1}4\right)\tag{4c}\\
&=\textstyle2\sqrt2\sqrt{t-1}+\frac{\sqrt2}2\sqrt{t-1}^3\tag{4d}
\end{align}
$$
Therefore,
$$
\begin{align}
\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)
&\approx2\sqrt2\int_1^x\sqrt{t-1}\,\mathrm{d}t+\frac{\sqrt2}2\int_1^x\sqrt{t-1}^3\,\mathrm{d}t\tag5\\[6pt]
&=\textstyle\frac{4\sqrt2}3(x-1)^{3/2}+\frac{\sqrt2}5(x-1)^{5/2}\tag6
\end{align}
$$

Graphs

