# 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?

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This doesn't really answer your question, but (to me) making the substitution $u = \sqrt{2x - 1}$ seems more natural. The fact that the answer turns out to be $2\tan^{-1}{(\sqrt{2x-1})}$ this way looks like it might suggest a motivation for considering the triangle you mentioned, if you stared at it for long enough – Daniel Freedman Jan 4 '12 at 22:06
This does not answer your question, so it is left as a comment. There is an obvious substitution $u^2=2x-1$. Then $dx=u\,du$, and after hardly any work we arrive at $\int \frac{2\,du}{1+u^2}=2\arctan u +C$. – André Nicolas Jan 4 '12 at 22:07
I'm not sure this will answer you but the side lengths of your triangle appear in the solution using this substitution: If $x={1\over 1-u}$, then $dx={ 1\over (1-u)^2}du$ and the integral becomes \eqalign{ \int{dx\over x\sqrt{2x-1}} &=\int {1-u\over \sqrt{2{1\over 1-u}-1 } } { 1\over( 1-u)^2}\,du\cr &=\int {1 \over \sqrt{2{1\over 1-u}-1 } } { 1\over 1-u}\,du\cr &=\int {1 \over \sqrt{{1+u\over1-u} } } { 1\over 1-u}\,du\cr &= \int {1 \over \sqrt{{1-u^2} } }\,du\cr &=\sin^{-1} u +C\cr &=\sin^{-1}{x-1\over x}+C. } – David Mitra Jan 4 '12 at 22:44
I suppose I should have been more clear: for motivation I was hoping for "this substitution makes sense for this particular class of ODEs," "this integral form has a general solution of 'such and such' and, in general, we use this substitution" or even my greatest hope "this substitution is a LF/Mobius transformation and it applies to this scenario because of ..." – NoClue Jan 5 '12 at 13:53
In any 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). – NoClue Jan 5 '12 at 13:55

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. – Gerry Myerson Jan 5 '12 at 1:17
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 $$\sqrt{2x-1}=\tan\beta=\cot(\pi/2-\beta)=\cot\angle OCA.$$