What is $e^{\int{\frac{1}{x}dx}}$? When doing integrating factors in order to solve differential equations, we often have something in a form similar to $$e^{\int{\frac{1}{x}dx}}$$
I know that $\int{\frac{1}{x}dx} = \ln{|x|}$, but for some reason, I often times see people simplifying $e^{\int{\frac{1}{x}dx}}$ down to just $x$.
Is $e^{\int{\frac{1}{x}dx}}$ equal to $|x|$ or $x$?
 A: In the most general sense
$$ \exp \left( \int \frac{1}{x}dx \right) = \exp(\ln|x| + c) = \pm e^C|x| = Cx $$
where $C$ can be positive or negetive
However, in the context of solving a linear first-order differential equation, neither the constant or the absolute value matters with respect to the integration factor. Since you're multiplying through the entire equation, both the constant and the sign will factor out.
Using your example, the equation
$$ y' + \frac{1}{x}y = f(x) $$
Suppose the integrating factor is $|x|$, you have two choices. Multiplying through by $x$ gives
$$ xy' + y = x f(x) $$
Multiplying through by $-x$ gives
$$ -xy' - y = -x f(x) $$
Both are obviously equivalent. Now you see why we drop the absolute value sign.

Note: If you're curious to why $1/x$ integrates to $\ln|x|$, the general anti-derivative is actually supposed to be
\begin{cases} \ln(x) + c_1, & x > 0 \\ \ln(-x) + c_2, & x < 0 \end{cases}
where you take either or both of the separate functions, depending on where the initial value is given. The two constants $c_1$, $c_2$ need not be the same.
A: In general, $$e^{\int\frac{1}{x}dx} = e^{\ln |x|} = |x|$$ is the correct answer. But often times, introductory classses on ODE deals with initial value problems and many of those problems have initial value given like $y(x_0) = y_0$, where $x_0\geq 0.$ In that case, you are only concerned with $x\geq x_0\geq 0$, so you can just forget about the absolute value. 
A: It is $|x|$. Notice that $e$ raised to a any real constant power $\ln|c|$ will always be positive, but $c$ may not necessarily positive due to the absolute value in the natural log. Take, for example:
$$e^{\ln|-1|} = e^0 = 1 \;\;\; -1 \neq 1$$
Remember that for all real constants $x$, $e^{\ln{x}} = x$, so $e^{\ln|x|}=|x|$.
A: It is $e^{\ln\vert x\vert}$, as you say (Notice that it is not defined at $x\neq0$), Since $\forall x\in\mathbb R^+,e^{\ln(x)}=x$, so it is$$e^{\ln\vert x\vert}=\vert x\vert$$
