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**Ackbeet** Well, when you integrate 0, you get a constant. You can call it whatever you want. In this situation, whoever is doing the integration must have had a reason for not wanting the constant to be negative (in this case, it's probably the square root that occurs later, thus avoiding complex numbers). Although, come to think of it, in quantum mechanics, complex numbers are common enough. I would need more context to be able to say why A must be real.

Yep. I was able to figure that out due to context.

Actually, in order for you to have a total differential, you should have this original expression:

$\displaystyle 2A(x)A'(x)\phi'(x)+A^{2}(x)\phi''(x)=0.$

If you start with this expression instead of your original expression, integration by parts works just fine. You'll get a cancellation that leads you to the same equation as above. Canceling an $\displaystyle A$ gives you your original expression.

Cheers.