TITLE: The Absolute Value in the Integral of $1/x$
QUESTION [39 upvotes]: $$\int\frac{1}{x}dx=\ln| x |+C$$
Why the absolute value? Why is the following not valid:
$$\int\frac{1}{x}dx=\ln x+C$$

REPLY [2 votes]: 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.).