When solving $3xy'+y=12x$, why can absolute sign in ln |x| can be ignored? So the following is the differential equation:
$3xy'+y=12x$
We can solve this by
recasting the equation to $y'+y/3x=4$
and finding an integration factor.
Integration factor is: $e^{\int 1/3x \,\, dx} = e^{1/3 \ln |x|} = |x|^{1/3}$
But often it seems that absolute signs are ignored. Why is it the case? Does the result not change whether absolute sign is there or not?
 A: Depending where we are in the domain, the expression $|x|$ either means $x$ or $-x$. If we keep the absolute value sign, we're really just saying the expression in question is one thing for $x>0$ and another for $x<0$. The $x>0$ case is what you get when you just ignore the absolute value sign.
If you write out the $x<0$ case - which I encourage you to do - you'll do the same calculation, but using $-x^{1/3}$ as your integrating factor. You'll see that the minus sign immediately drops out, so it doesn't make any difference at all. That's why we don't bother with it.
A: The modulus signs are just in there to ensure that $\ln$ has a positive argument.  The general antiderivative of $\frac1x$ is $\ln(x)+c$.  Notice that $\ln(kx)=\ln(x)+\ln(k)$ so, since indefinite integrals are only defined up to an arbitrary constant, $\int \frac1xdx$ can take the value $\ln(kx)$ for any $k$.  In particular, since we don't normally want to take the logarithm of a negative number, we can set $\int\frac1xdx=\ln(x)+c$ for positive $x$ and $\int\frac1xdx=\ln(-x)+c$ for negative $x$.  This isn't a problem, since $\frac1x$ isn't defined at $0$ anyway, so we are never going to try to integrate $\frac1x$ over an interval that includes the point $0$.  
If you are given the definite integral $\int_{-2}^{-1}\frac1xdx$, then you can use the antiderivative $\ln(|x|)$ to give you the answer $\ln(1)-\ln(2)=-\ln(2)$.  If the antiderivative were just $\ln(x)$ then you would run into difficulty trying to work out $\ln(-1)$.  
When you try to find an integrating factor in your example, you are looking for $e^{F(x)}$, where $F(x)$ is an antiderivative for $\frac1{3x}$; i.e., $\frac{dF}{dx}=\frac1{3x}$.  Now, you could take $F(x)=\frac13\ln(|x|)$, and that would be a perfectly fine antiderivative.  However, taking $F(x)=\frac13\ln(x)$ gives you $x^{1/3}$, which is a bit easier to deal with.  Of course, if the exponent were $1/2$ rather than $1/3$ then you would want to use $|x|^{1/2}$ since you can't take the square root of a real number and end up with a real number.  
Now, I'm sure you know that we can't take the logarithm of a negative number.  Why, then, is it OK to say our antiderivative $F$ is $\frac13\ln(x)$ for both positive and negative $x$?  One way we could deal with this problem is just to ignore the fact that we're doing something forbidden and just notice that it all ends up fine in the end when we pass to $e^{F(x)}$.  Another, slightly more enterprising, way to tackle the problem is just to pretend that $\ln(-1)$ has a value, and then notice the fact that $\ln(-x)=\ln(x.-1)=\ln(-1)$.  Thus we can now define $\ln$ of a negative number in terms of our new value $\ln(-1)$.  We can even do the integral $\int_{-2}^{-1}\frac1xdx$ without resorting to the messy $\ln(|x|)$ antiderivative: the integral is equal to $\ln(-1)-\ln(-2)=\ln(-1)-\ln(-1)-\ln(2)=-\ln(2)$.  Then our magical $\ln(-1)$ value disappears when we pass it through the exponential function: obviously, $e^{\ln(-1)}=-1$, and we are back to numbers that we are familiar with and didn't have to make up.  
In fact, we can do even better that.  Perhaps you know the famous identity
$$
e^{i\pi}=-1
$$
where $i$ is the square root of $-1$ (which was itself originally invented in order to solve problems involving real numbers: terms involving $i$ appeared in the middle of calculations, but they had all disappeared by the end).  This means that we can take $\ln(-1)=i\pi$, and hence that $\ln(-x)=i\pi+\ln(x)$.  So, if we move to the complex numbers, we can do without the $\ln(|x|)$ construct altogether.  
[In fact, there are some problems even with this approach, because we have to be careful in our definition of $\ln(-1)$.  For example, $e^{3i\pi}=-1$ as well, so why can't we take $\ln(-1)=3i\pi$?  In fact, this value works as well, but we have to choose one value (called choosing a branch of the logarithm and stick with that throughout.  That's fine for solving simple question like this one, though.  This seeming problem actually leads to some remarkable and beautiful mathematics in the field of complex analysis - in particular, the study of Riemann surfaces.]
