The Absolute Value in the Integral of $1/x$ $$\int\frac{1}{x}dx=\ln| x |+C$$
Why the absolute value? Why is the following not valid:
$$\int\frac{1}{x}dx=\ln x+C$$
 A: Your range of integration can't include zero, or the integral will be undefined by most of the standard ways of defining integrals.  So we have to think of a range of integration which is strictly positive, or strictly negative.
What you wrote is perfectly valid for strictly positive x, so let's think about strictly negative x.  We have
$\int_{-a}^{-b}\frac{1}{x}d x$
where $a>0$ and $b>0$, so the range of integration is strictly negative.  Do a change of variables, $y=-x$.  Then
$\int_{a}^{b}\frac{1}{y}d y$.
(There is a negative from the $y$ in the denominator, and $d x=-d y$, so the two negatives cancel.)  We have converted the integral of $1/x$ over a strictly negative range to an integral of $1/y$ over a strictly positive range.  The answer is $\ln b-\ln a$. Since the $y$ is just a variable of integration, we can replace it with $x$ if we like, and
$\int_{-a}^{-b}\frac{1}{x}d x=\int_{a}^{b}\frac{1}{x}d x$.
That's the definite integral; the analogous result for the indefinite integral is
$\int^{-x}\frac{1}{x}d x=\int^{x}\frac{1}{x}d x$ (to within a constant of integration).
A: For $x$ positive:
$
\frac{d}{dx}\ln{x}=\frac{1}{x}
$
For $x$ negative:
$
\frac{d}{dx}\ln{(-x)}=\frac{-1}{-x}=\frac{1}{x}
$
So when you're integrating $\frac{1}{x}$, if $x$ is positive you'll get $\ln{x}+C$, and if $x$ is negative you'll get $\ln{(-x)}+C$. To summarize $\ln{|x|} + C$.
And if you want to know $\int\frac{1}{x}dx$ is not exactly equal to $\ln|x|+C$. The constants could be different for positive or negative $x$.
$$
\int\frac{1}{x}dx =
\begin{cases}
\ln{x} + C_1 \qquad \text{for $x$ positive} \\
\ln{(-x)} + C_2 \qquad \text{for $x$ negative}
\end{cases}
$$
A: I will offer a very simple intuitive approach.
If we take:
$$ln(x)=\int_{1}^x \frac1u du$$
We find that $u=0$ is a point of discontinuity for the function $1/u$. So, you might notice that, for example, that for $ln(x)$, as $x$ approaches $0$, the function $ln(x)$ increases "beyond all bounds" in the negative direction, or that $ln(1)=0$:
$$ln(1)=\int_{1}^1 \frac1u du=0$$

Supplement - If we take $F(x)$ as any primitive function and keeping in mind of the fundamental theorem of calculus where any primitive function differ only by a constant (in this case, the indefinite integral(s) of the natural logarithm, differing by a constant, makes intuitive sense) such that:
$$F(x)=ln(x)+c=\int_{1}^x \frac1u du +c$$
$$\frac{d}{dx}F(x)=\frac{d}{dx}(ln(x)+c)=1/x$$
To be specific, the definition of the primitive function here is merely just $\frac{d}{dx}F(x)=f(x)$.

This answer is just to offer some basic intuitive sprinkle on the agreeable assertion "Your range of integration can't include zero..."@Роберт
Further Supplement - For further basic intuition, notice that $1/u$ as defined for the domain $(0,1]$ lacks a uniform modulus of continuity. However, the existence of a uniform modulus of continuity is implicit in the existence of any integral (as suggested by e.g. Bolzano–Weierstrass theorem etc., and in where I would say as according to the traditional definition of an integral as suggested by e.g. $\lim \limits_{\Delta x \to 0}\sum_{i}={f(x_i)}{\Delta x}$ defined for the interval $[a,b]$ such that $a+0\Delta x=a$ and $a+n\Delta x=b$ etc.). 
A: $\int_{-2}^{-1}\frac{1}{x}dx$ has a value, however $\ln(-1)$ and $\ln(-2)$ will be more complicated to evaluate since $\ln(x)$ is only defined on $\mathbb{R}$ for positive numbers...
Actually since $\frac{1}{x} = -\frac{1}{-x}$ for every $x$, we have
$$\ln(|-1|)-\ln(|-2|)=\int_{-2}^{-1}\frac{1}{x}dx=\int_2^1\frac{1}{x}dx = \ln(1)-\ln(2) = \ln\left(\frac{1}{2}\right)<0.$$
A: The function $f(x)=\frac{1}{x}$ is not integrable on any interval that contains $x=0$. Therefore, any integral of this function is on an interval that is either entirely positive or entirely negative.
Let $[a,b]$ be such an interval.
Case 1: $0<a<b$
$$\int_a^b \frac{1}{x}dx=\log(b)-\log(a)$$
This is by definition, since the natural logarithm is defined as
$$\log(x)=\int_1^x \frac{1}{x}dx, \text{ for }x>0$$
hence
$$\int_a^b \frac{1}{x}dx=\int_1^b \frac{1}{x}dx - \int_1^a \frac{1}{x}dx=\log(b)-\log(a)$$
Now, there are of course infinitely many other functions whose derivatives are $\frac{1}{x}$. By the first fundamental theorem of calculus, since, $\frac{1}{x}$ is continuous on $(0,\infty)$, any function $F(x)=\int_{x_0}^x \frac{1}{x}dx$ with $0<x_0<x$ is such that $F'(x)=\frac{1}{x}$.
Such a function can be rewritten as $$F(x)=\int_1^x \frac{1}{x}dx-\int_1^{x_0} \frac{1}{x}dx$$
$$=\log(x)-\log(x_0)$$
Hence, for any of such functions we still have
$$\int_a^b\frac{1}{x}dx=(\log(b)-\log(x_0))-(\log(a)-\log(x_0))$$
$$=\log(b)-\log(a)$$
$$=\log(|b|)-\log(|a|)$$
Now we have the tricky case.
Case 2: $a<b<0$
We want $\int_a^b\frac{1}{x}dx$.
Let $f(x)=\frac{1}{x}$, and here is its graph

The quick argument here is that, by symmetry
$$\int_a^b \frac{1}{x}=-\int_{-b}^{-a}\frac{1}{x}\tag{1}$$
and hence, since the right side involves a positive interval, we know how to compute it
$$=-(\log{(-a)}-\log{(-b)})=\log(|b|)-\log(|a|)$$
Thus in both cases, we have
$$\int_a^b \frac{1}{x}=\log(|b|)-\log(|a|)$$
Addendum
What does "by symmetry" mean more precisely in that first step (1)?
As far as I can tell, it boils down to an argument based on the definition of the integrals involved. Such an argument is not necessary to understand the symmetry, since it is quite obvious, but to understand what is going on under the hood I think the proof of the symmetry would involve going through calculations such as the following.
Let $P=\{a=t_0,t_1,...,t_n=b\}$ be a partition of $[a,b]$ and $Q=\{-b=-t_n, -t_{n-1},...,-t_0=-a \}$ be a partition of $[-b,-a]$. Then
$$L(f,P)=\sum\limits_{i=1}^n \frac{1}{t_i}(t_i-t_{i-1})=\sum\limits_{i=1}^n \frac{1}{-t_i}(-t_{i-1}-(-t_i))$$
$$=\sum\limits_{i=1}^n \frac{1}{-t_i}(t_i-t_{i-1})=-U(f,Q)\tag{2}$$
$$U(f,P)=\sum\limits_{i=1}^n \frac{1}{t_{i-1}}(t_i-t_{i-1})=\sum\limits_{i=1}^n \frac{1}{-t_{i-1}}(-t_{i-1}-(-t_i))$$
$$=\sum\limits_{i=1}^n \frac{1}{-t_{i-1}}(t_i-t_{i-1})=-L(f,Q)\tag{3}$$
where $L(f,P)$ is the lower sum of $f$ for partition $P$, and $U(f,P)$ is the upper sum of $f$ for partition $P$.
By definition $\int_a^b f=\sup{L(f,P)}=\inf{U(f,P)}$ and $\int_{-b}^{-a} f=\sup{L(f,Q)}=\inf{U(f,Q)}$
But from (2) and (3) we also have that $\sup{L(f,P)}=\sup{[-U(f,Q)]}=-\inf{U(f,Q)}$ and $\inf{U(f,P)}=\inf{[-L(f,Q)]}=-\sup{L(f,Q)}$
Thus
$$\int_a^b f=-\int_{-b}^{-a}f$$
