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?