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I was working through a problem I invented while devising homework problems for my thermodynamics course. I encountered a crazy integral and was able to show using conservation of energy that the integral has a very simple solution. I checked the answer using numerical integration and it holds to arbitrary accuracy.

The integral $I$ is $$ I = \int_1^{x'} \frac{df(x)}{x} $$ where $$ x' = \frac{\beta}{\beta+1} $$ with $\beta > 0$ and $$ f(x)=\frac{(3-y)y^{1/2}}{(2-y)^{3/2}} $$ where $$ y(x)=1 - \sqrt{1 - \left(\beta \times \frac{1-x}{x}\right)^2} $$ I find that $$ I = 2 + \frac{3}{2} \times \frac{1}{\beta} $$ How can such a complicated looking integral have such a simple solution? The problem is essentially a calculation of the work required to inflate a soap bubble.

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    $\begingroup$ Also, if you have a solution via conservation of energy then it would seem worth including as its own answer. $\endgroup$ Commented Jun 16 at 2:16

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This is not a complete solution---see velut luna's answer for the calculation of one of the definite integrals---but explains the behavior seen.

Note that $x\mapsto y(x)$ maps $[\beta/(\beta+1),1]$ to $[0,1]$, so the integral can be expressed in terms of $y$: $$I=\int_0^1 \frac{df(y)}{x(y)}$$ Moreover, it's not too bad to solve for $x$ in terms of $y$:

\begin{align} y&=1-\sqrt{1-\left(\beta\frac{1-x}{x}\right)^2}\\ 1-(1-y)^2&=\left(\beta\frac{1-x}{x}\right)^2\\ 1-\frac{1}{x}&=-\beta^{-1}\sqrt{1-(1-y)^2}\\ \implies x(y)&=\left[1+\beta^{-1}\sqrt{2y-y^2}\right]^{-1} \end{align} where the minus sign of the square root is taken to recover $x(1)=\beta/(\beta+1)$. Hence the original integral can be expressed as

$$I=\int_0^1 \left(1+\beta^{-1}\sqrt{2y-y^2}\right)~df(y)$$

This is now evidently of the form $I=I_0+\beta^{-1} I_1$ for appropriate $\beta$-independent definite integrals $I_0,I_1$, verifying the functional form observed by the OP. The first can be immediately integrated to obtain $I_0=\int_0^1 df(y)=f(1)-f(0)=2.$ The second is not so obvious and I don't have a satisfactory solution. But according to Mathematica the integrand can be expressed as

$$\sqrt{2y-y^2}\frac{df}{dy}=\frac{d}{dy}\left(\frac{3}{2-y}\right)$$ and hence $I_2=3/(2-1)-3/(2-0)=3/2$. This verifies $I=2+\frac{3}{2\beta}$.

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Following Semiclassical's answer, the second integral is not difficult as well. $$\int_0^1\sqrt{2y-y^2}df(y)=\sqrt{2y-y^2}f(y)\bigg|_0^1-\int_0^1f(y)d\left(\sqrt{2y-y^2}\right)$$ $$=2-\int_0^1\frac{(3-y)\sqrt{y}}{(2-y)^{3/2}}\frac{2-2y}{2\sqrt{2y-y^2}}dy$$ $$=2-\int_0^1\frac{(3-y)(1-y)}{(2-y)^2}dy$$ $$=2-\int_1^2\frac{(u+1)(u-1)}{u^2}du$$ $$=2-\int_1^2\left(1-\frac{1}{u^2}\right)du$$ $$=2-\left(u+\frac{1}{u}\right)\bigg|_1^2$$ $$=2-1+\frac{1}{2}$$ $$=\frac{3}{2}$$

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