Integrating factor with $-\ln x$ I know that
$$\int \frac{1}{x}dx = \ln |x| + C$$
however I have seen differential equation notes and solutions claim that the integrating factor for $P(x)=-\frac{1}{x}$ is
$$\mu(x)=e^{\int P(x)dx}=e^{-\int\frac{1}{x}dx}=e^{-\ln|x|}=\frac{1}{x}$$
For example consider the IVP
$$\frac{dy}{dx}-\frac{y}{x}=xe^x, ~~~y(1)=e^{-1}$$
We have that $P(x)=-\frac{1}{x}$ so we could find the integrating factor exactly as above
$$\mu(x)=e^{\int P(x)dx}=e^{-\int\frac{1}{x}dx}=e^{-\ln|x|}=\frac{1}{x}\tag{1a}$$
then our equation would become
$$\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=e^x \implies \frac{d}{dx}\Big(\frac{1}{x}y\Big)=e^x$$
which after integrating produces
$$\frac{y}{x}=e^x+C$$
Applying the initial condition of $y(1)=e^{-1}$ forms $C=-1$. Then
$$\frac{y}{x}=e^x+1$$
or
$$y=xe^x+x\tag{1b}$$
If instead we found the integrating factor as
$$\mu(x)=e^{\int P(x)dx}=e^{-\int\frac{1}{x}dx}=e^{-\ln|x|}=\frac{1}{|x|}\tag{2a}$$
then we would carry through the $|x|$ throughout the computation. We have
$$\frac{1}{|x|}\frac{dy}{dx}-\frac{y}{x|x|}=e^x \implies \frac{d}{dx}\Big(\frac{1}{|x|}y\Big)=e^x$$
which after integrating forms
$$\frac{y}{|x|}=e^x+C$$
Applying the initial condition of $y(1)=e^{-1}$ once again forms $C=-1$. Then
$$\frac{y}{|x|}=e^x+1$$
or
$$y=|x|e^x+|x|\tag{2b}$$
I have seen different people claim that both solutions are correct. I'm not sure if we can drop the absolute value sign at some point in the computation. 
 A: The differential equation breaks down at $x=0$, so what happens for negative $x$ is something we cannot tell from the given information. We only care about positive $x$ because we can only care about positive $x$, and therefore the absolute value signs do nothing.
Even for the simpler differential equation $y'=\frac yx$, we get general solution
$$
y(x)=\cases{ax& for $x>0$\\bx& for $x<0$}
$$
In fact, the true antiderivative of $\frac1x$ is
$$
\cases{\ln x+c_1& for $x>0$\\\ln(-x)+c_2 & for $x<0$}
$$
and you don't have any information to help you pin down $c_2$ (or really $c_2-c_1$) for your integrating factor, which is another manifestation of our inability to tell what happens for negative $x$.
Of course, assuming something like $y$ being differentiable at $x=0$ (if that's even something that can happen; that's not the case with all differential equations) will be enough to stitch together the negative and positive branches.
A: We can get the integrating factor without logarithms.  Use the quotient rule for differentiation:
$\dfrac{(u/v)}{dx}=\dfrac{v(du/dx)-u(dv/dx)}{v^2}$
Put $u=y, v=x$ to get
$\dfrac{d(y/x)}{dx}=\dfrac{x(dy/dx)-y(dx/dx)}{x^2}=\dfrac{1}{x}\dfrac{d(y/x)}{dx}$
which gives the integrating factor and the exact integral in one blow.  Since this applies for both positive and negative $x$ we have the $(1b)$ solution unambiguously.
