Why is $\int \frac{f'(x)}{f(x)}=\ln |f(x)|$ ignored in differential equations?
Review of Integrating Factors
Consider integrating factors in the context of linear equations, say $\dfrac{\mathrm{d}y}{\mathrm{d}t}+p\left(t\right)y=g\left(t\right)$. (I'll often drop the input $t$s from now on.) Our hope is to multiply the equation by some function $\mu$ to turn the left side into the derivative of a product, say $\left(\mu y\right)'$, so that we can integrate both sides and divide by $\mu$ to solve for $y$. Note that by the product rule, we have $\left(\mu y\right)'=\mu'y+\mu y'$. To get $\left(\mu y\right)'=\mu*\left(y'+py\right)$, we should look for $\mu$ that satisfy $\mu'=\mu p$. (That's a separable equation.)
Absolute values in integrating factors mostly don't matter
Note that if $\mu\left(t\right)$ is a valid integrating factor (in the sense that $\mu'\left(t\right)=\mu\left(t\right)p\left(t\right))$, then define $\nu(t)=-\mu(t)$ and note that $\nu'\left(t\right)=-\mu'\left(t\right)=-\mu\left(t\right)p\left(t\right)=\nu\left(t\right)p\left(t\right)$. But $\nu'=\nu p$ is exactly the equation $\nu$ should satisfy to be an integrating factor! So the negative of an integrating factor works just as well.
Now suppose you solve $\mu'=\mu p$ and get a solution like $\mu\left(t\right)=\left|\omega\left(t\right)\right|$, thanks to an integral of $\dfrac{\omega'\left(t\right)}{\omega\left(t\right)}$. On an interval where $\omega\left(t\right)$ is always nonnegative, $\mu\left(t\right)=\omega\left(t\right)$, and you don't need the absolute value. On an interval where $\omega\left(t\right)$ is always nonpositive, $\mu\left(t\right)=-\omega\left(t\right)$, so $\nu\left(t\right)=-\mu\left(t\right)=-\left(-\omega\left(t\right)\right)=\omega\left(t\right)$, and you still don't need to worry about the absolute value, since $\nu$ works fine, too.
Therefore, if you ignore the absolute value, you get valid solutions on these intervals where $\omega$ doesn't change sign.
In practice, absolute values in integrating factors don't matter at all
A limitation of the above discussion is that there's no guarantee the solutions on separate intervals will be able to connect up when $\omega\left(t\right)=0$. However, that's okay, because in practice an issue like that would usually arise because the original differential equation has a problem at that point.
For example, with something like $ty'+3y=te^{t}$, you would rewrite as $y'+\dfrac{3}{t}y=e^{t}$ and initially get an integrating factor of $\left|t\right|^{3}$. By the above argument, for positive $t$ or negative $t$, you can get away with $t^{3}$ and learn that $t^{3}y=\int_{0}^{t}e^{x}\,\mathrm{dx}+C$. But what about at $t=0$? Well, that would make $t^3y$ not tell you about $y$, it would make $\dfrac{3}{t}$ undefined, and would make $ty'+3y=te^{t}$ not tell you anything about $y'$. In fact, the solutions for positive or negative $t$ diverge as $t$ approaches $0$, so you had no hope of finding a solution on an interval like $\left[-1,1\right]$ anyway.