Integration using $x = 2\cosh u$

I'm working on the problems in this booklet: https://mcs-notes2.open.ac.uk/files/MScDiagnosticquiz.pdf

In question 1.2.1(f) the integration is:

$$\int_{2}^{3}\frac{x+1}{\sqrt{x^2-4}} \,dx$$

Later on in the booklet (Page $18$) there is a solution which I understand fully (cannot integrate between $2$ and $3$ in its current form, use a substitution, integrate that and use the new form which is continuous at $x = 2$. That's all good.

What I'm struggling with though is - why $ x = 2\cosh u $? Although I can follow the solution in full, I don't know where the substitution came from or how I would know to use that substitution in a similar question.

You have $\cosh^2-1=\sinh^2$. So, the function $\cosh$ has this property: its square minus $1$ is the square of a known function. More generally, if $k\in\Bbb R$, the square of $k\cosh$ minus $k^2$ is the square of a known function ($k\sinh$). Therefore, if you do $x=2\cosh u$, then $\sqrt{x^2-4}$ becomes $2\sinh u$.

The same thing occurs with the secant function, since $\sec^2-1=\tan^2$,