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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?

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    I$n$ a$n$y event, David Mitra appears to have the closest approximation thus far regarding "inspiration" in that the argument of the arcsin contains the substitution (interestingly, one can also get a solution in arctan -- the integration constant is the medium of equality between the two solutions).2012-01-05

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Sometimes, the "motivation" for a substitution is you've solved the problem some other way and you look at the solution and you notice that there was this substitution that would have made everything work out nicely.

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    Cut me a little slack here. The answer comes out to be $\arcsin((x-1)/x)+C$, according to David Mitra's calculations. You look at that and say, why didn't I substitute $u=(x-1)/x$ in the first place? Well, $u=(x-1)/x$ is $x=1/(1-u)$. The answer "motivates" the substitution.2012-01-05
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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. $