What is the Simplified Integral for Average Momentum?

eck
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I have the physics question along with the solution, but in the solution I don't understand how they evaluated the integral. I can't get my brower to preview TeX input, so I'm going to leave it without formatting, but you can find the problem http://web.mit.edu/8.05/probsets/ps1_v1.pdf". The problem I am looking at is number one. I've also got the problem and solution posted here (w/o formatting) but it's kind of hard to read:
Problem
--------------------
A particle's coordinate space wavefunction is square-integrable and real up to an arbitrary multiplicative phase:
psi(x) = exp(i * alpha) phi (x)
with alpha real and constant and phi(x) real. Prove that its average momentum is zero.
Solution
-------------------
Setting up the integral is easy, and you can pull out a couple constants. So you have an infinite integral with this inside:
dx exp(-i alpha) phi(x) exp(i alpha) d/dx[phi(x)]
Somehow, in the solution, they pull out 1/2 and leave the following in the integral:
dx d/dx[phi(x)^2]
When I look at it, I see the exponentials cancelling, but I don't understand where the 1/2 comes from and how the first phi(x) gets pulled into the derivative.
Can anyone shed any insight on how this integral was simplified?
 
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I was looking at the problem some more, and all of a sudden it hit me. It's kind of embarrassing that I didn't see it before. If anyone else looks at it... nothing tricky is involved. It's integration by parts, but it's so obvious I didn't even see it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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