Motivation for a particular integration substitution In an old Italian calculus problem book, there is an example presented:
$$\int\frac{dx}{x\sqrt{2x-1}}$$
The solution given uses the strange substitution $$x=\frac{1}{1-u}$$
Some preliminary work in trying to determine the motivation as to why one would come up with such an odd substitution yielded a right triangle with hypotenuse $x$ and leg $x-1;$ determining the other leg gives $\sqrt{2x-1}.$  Conveniently, this triangle contains all of the "important" parts of our integrand, except in a non-convenient manner.
So, my question is two-fold:
(1)  Does anyone see why one would be motivated to make such a substitution?
(2)  Does anyone see how to extend the work involving the right triangle to get at the solution?

This won't answer the question, but it takes the geometry a bit beyond where the question left it.  Consider the circle of unit radius in the Cartesian plane centered at $O=(0,1)$.  Let $A=(1,0)$  and $B=(x,0)$.  Let $C=(1,\sqrt{2x-1})$.  Your right triangle is $ABC$, with angle $\alpha$ at vertex $B$.  Another right triangle is $OAC$, with angle $\beta$ at $O$.  Then
$$
u=\frac{x-1}{x}= \cos\alpha=\sin(\pi/2-\alpha)=\sin\angle BCA$$
and
$$
\sqrt{2x-1}=\tan\beta=\cot(\pi/2-\beta)=\cot\angle OCA.
$$
