Is the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$? The inconsistency I see between mathematical subjects is really confusing me. I understand that it isn't possible for $e^x$ to be less than zero for real $x$, which is probably why they say that the integral is $\ln(|x|)$.
Before I ramble on too much, I just want to ask: Is there a set of guidelines to follow to help me choose whether to let the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$?
Thanks,
Aralox
 A: In general, it is safe to always write:
$$\int \frac{1}{u} \ du = \ln \left| u \right| +C $$ where $C$ is some constant.
However, if the function is always positive $\forall x  \in \Bbb R\setminus \{0\}$ (assuming that's what you're integrating over), you can drop the absolute value. 
Addendum of Complex Logarithm
I think this will be helpful since you wanted to know how to integrate with the complex log. Also, this should be a good read. I'd start around page 74 for what you're looking for.
I feel it might help to note that if we say $\log z$ is any logarithm along some branch $B$, then $(\log z)' = \dfrac{1}{z} \forall z$ not on $B.$ However, no matter how we define the complex logarithm, there will always be some branch that is not holomorphic.
A: Equal to $\ln(|x|)$+c, in which $c$ is constant.
A: The correct answer is$\int \frac1x dx= \ln  |x| +C$
The absolute value is sometimes omitted in ODE problems. As for guidelines I would say analyze the problem and see if values of x will be out of the domain when solved. 
A: Strictly speaking, any function of the form
$$F(x) = \begin{cases}
\ln x + c_1 & \text{if } x > 0, \\
\ln (-x) + c_2 & \text{if } x <0
\end{cases}$$
defined over $\mathbb R \setminus \{0\}$ is a valid antiderivative. Verify this by differentiating $F(x)$ and getting $F'(x) = 1/x$ back for any $x \ne 0$.
The usual formula, $\ln\lvert x\rvert + c$, is what you get when you pick $c_1 = c_2 = c$. Alternatively, as @Joe says, when you're only considering positive $x$, you can just write $\ln x + c$.
A: I think the introduction of complex variables here is unnecessary. The function $1/x$ is a perfectly well-defined function on $\mathbb{R} \setminus \{0\}$ and we can ask if it has an antiderivative on that set. On the positive real axis, $\log(x)$ is an antiderivative. On the negative real axis, $\log(-x)$ is an antiderivative (and $\log(x)$ doesn't make sense). This is confusing, so we write $\log(|x|)$ so that we don't have to remember. Note that $\log(|x|) + C$ is not good notation, since the constant can be different on the negative and positive real axes.
A: 
I remember studying the principal logarithm a few months back, and
  understand that Ln(z) = ln|z| + iArg(z). Could you show me how to
  integrate 1/x using it?

Right, and this is why the "most" correct answer is that $\ln(x)$ would be the antiderivative, considered as a function over the complex numbers (still not defined at zero).  So that (as an example) one could use $\ln(-x)=\ln(x)+i\pi$ in the manner of Rahul's answer (with $c_2=i\pi$, which Euler wonderfully treats as a constant to ignore in some of his papers).  
This has the problem that it's a multivalued function, but has the advantage that this enables one to choose a consistent branch to integrate over all sorts of interesting contours.  
For instance, then the integral of $1/x$ over the unit circle (in $\mathbb{C}$) has to follow from $\ln(1)=0$ to $\ln(1)=2\pi i$ giving $2\pi i-0 = 2\pi i$, where I've here really followed the function up the Riemann surface, as it were - obviously not a function any more.  Not how we usually think of the FTC, granted!   But that really is the value of the contour integral.
See http://www.ma.utexas.edu/maxima/maxima_13.html for one CAS' solution to this issue.
A: On the negative axis their difference is a constant. Both are antiderivatives. 
