Why is $\int \frac{f'(x)}{f(x)}=\ln |f(x)|$ ignored in differential equations? I've noticed that whenever $\int \frac{f'(x)}{f(x)}$ comes up in a differential equation the answer is always given as $\ln f(x)$ rather than $\ln |f(x)|$ as I was taught it should be. Is it because of the arbitrary constant? In other words, since $$\int \frac{f'(x)}{f(x)}=\ln |f(x)|+\ln A$$ for some constant $A$, then the answer is $\ln A|f(x)|$ and because $A$ can be positive or negative it follows that there is no point including the absolute signs? Hence the answer is given as $\ln f(x)+C$ for some constant $C$ rather than $\ln |f(x)|+C$. Is this why?
 A: 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.
A: Writing $ \int \frac{f'(x)}{f(x)} = \ln f(x)$ implicitly supposes that $f(x) > 0$. The logarithm is not defined for negative numbers.

Why isn't the general formula needed?
Consider the differential equation
$$
y' = ty.
$$
To find the solutions to such equations, we can proceed in two steps.
Step 1. Suppose we have a solution $y: I \rightarrow \mathbb{R}$, where $I$ is an open interval. Further assume $y(t) \not = 0$, for all $t$. Since $I$ is an interval, then either $y > 0$ or $y < 0$. In the first case, we find $\frac{y'(t)}{y(t)} = t$, from which $\ln y(t) = t^2/2 + C$ and $y(t) = \alpha e^{t^2/2}$, for some $\alpha > 0$.
Step 2. We verify that $y(t) = \alpha e^{t^2/2}$, $I = \mathbb{R}$, is indeed a solution for all $\alpha \in \mathbb{R}$ and that all solutions have been found.
As you can see, when everything is done so carefully, there is no need to use the general formula $\int \frac{f'}{f} = \ln|f|$. I guess this is what teachers have in mind when presenting an abbreviated argument and writing "$\int \frac{y'(t)}{y(t)} = \ln y(t)$".
A: Imho, the author is usually assuming $f$ takes positive values. This is reasonable considering many applications.
If $f$ does not necessarily take positive values, you still have a misunderstanding concerning the "constant" of integration. If by $\int g(x)\,dx$ you mean a fully generic antiderivative, then $$\int \frac{f'(x)}{f(x)}\,dx=\ln\left\lvert f(x)\right\rvert+C(x)=\ln\left\lvert e^{C(x)}f(x)\right\rvert=\ln\left\lvert A(x)f(x)\right\rvert$$ where $C$ is not exactly a constant function. $C$ is piecewise constant, allowed to change its value wherever $f(x)=0$. Consequently $A$ is also allowed to change value at places where $f(x)=0$.
Even with $\int\frac{1}{x}\,dx$, functions like $$\ln|x|+\begin{cases}1&x\gt0\\-1&x\lt0\end{cases}$$ are valid antiderivatives of $\frac{1}{x}$.
A: With separable equations, we're often solving for $y$ in the integral equation 
$$\int \frac{dy}{y} = \int p(x) \ dx$$
so that, given $P$ is an antiderivative of $p$, we have
$\ln |y| = P(x) + C_1$
where $-\infty < C_1 < \infty$. Hence, $\exp$ing both sides we get 
$$|y| = C_2\exp P(x)$$ where 
$0<C_2<\infty$
Note that $|y| \neq 0$ since the right hand side is never $0$. Therefore, $y(x)$ is either strictly positive or strictly negative and hence we can simply consider a new nonzero constant $C_3$ which ranges from $-\infty < C_3 < \infty$
and note that $y(x)$ satisfies
$y(x) = C_3 \exp P(x)$
We would have gotten the same result if we had ignored the absolute value. 
A: Mathematicians doing ODE usually assume complex number calculations.  In that setting, the assertion
$$
\int\frac{dx}{x} = \log |x| + C
\tag{1}$$
is simply wrong.  $\log |x|$ is not complex differentiable at any point.  In particular, it does not have derivative $1/x$.  
But, as you note, you can still write
$$
\int \frac{dx}{x} = \log x + C
\tag{2}$$
even when $x<0$.  The $\log$ and the $C$ may both be non-real complex numbers.  But only beginners would worry about that.  
I think formulas of the type (1) were invented by textbook writers within the last 50 years or so.
