What should $\int \frac{1}{x} dx$ equal to? Before you say that $\int \frac{1}{x} dx$ is equal to $\ln|x| +C$ due to positve and negative, I would like to show you why it is not convincing to me.

Problem 1 and its possible solution.
Evaluate
$$
\begin{equation} \sum_{n=1}^\infty \frac{\sin(n)}{n}
\end{equation}
$$
From infinite geometric series
\begin{equation}
\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x} ;|x|<1
\end{equation}
Integrating this with respect to $x$ we would get (Constant vanishes due to $x=0$)
\begin{equation}
\sum_{n=1}^\infty \frac{x^n}{n}=-\ln|1-x|
\end{equation}
So $\sum_{n=1}^\infty \frac{\sin(n)}{n}$ is just an imaginary part of
\begin{equation}
\sum_{n=1}^\infty \frac{z^n}{n}=-\ln|1-z|
\end{equation}
Where $z=e^i$
But the right hand side has no imaginary part at all, but the summation is clearly exists, this seems to suggest that the integral is equal to $\ln(x) +C$There is another curious way to evaluate integral on negative reals if we only consider only principal values.
Problem 2 and its possible solution. Evaluate
\begin{equation} \int_{-4}^{-2} \frac{1}{x} dx
\end{equation}
If we give that $\int \frac{1}{x} dx=\ln(x) +C $ then the integral is
\begin{equation} \int_{-4}^{-2} \frac{1}{x} dx=\ln(-2)-\ln(-4)
\end{equation}
Using principal values we will get
\begin{equation} \ln(-2)-\ln(-4)=\ln(2)+i\pi-\ln(4)-i\pi=-\ln(2)
\end{equation}
Which is exactly equal to when we use $\int \frac{1}{x} dx=\ln|x|+C$
 These 2 problems are the reasons why the result $\ln |x| +C$ not convincing but there might be flaws in the proposed solutions. If there is a flaws please explain them too.
 A: Without a clear definition of "$\log(x)$" and "$\int\frac{1}{x}\ dx$", it is meaningless to ask $\int\frac{1}{x}\ dx$ should be $\log(x)+C$ or $\log(|x|)+C$. 
On the one hand, the notation 
$$
\int f(x)\ dx\tag{0}
$$
(some times called the anti-derivative of $f$) really means (by definition) a function $F$ such that 
$$
F'(x)=f(x)\tag{1}
$$
where $x$ takes values so that $(1)$ is true. 
On the other hand, one way to define the natural logarithm is by Riemann integrals:
$$
\log(x)=\int_1^x\frac{1}{t}\ dt,\quad x>0.\tag{2}
$$
Whenever $a<b$, we use the convention that
$$
\int_b^af(x)\ dx=-\int_a^b f(x)\ dx.
$$
By the fundamental theorem of calculus and chain rule, 
$$
\frac{d}{dx}\log(-x)=\frac{1}{-x}\cdot (-1)=\frac{1}{x}
$$
for $x<0$ and
$$
\frac{d}{dx}\log(x)=\frac{1}{x},\quad x>0.
$$
In either case we have
$$
\frac{d}{dx}\log(|x|)=\frac{1}{x},\quad x\neq 0.\tag{3}
$$
Since two function have the same derivative if and only they are the same up to a constant (by the mean value theorem), we have
$$
\int\frac{1}{x}\ dx=\log(|x|)+C,\quad x\neq 0\tag{4}
$$
Note that until now, no complex numbers are evolved at all. The two problems in OP are irrelevant to $(4)$ because in complex analysis, the complex logarithm has totally different meaning from the real one (related though) and we are now in the setting of real numbers. 
A: The formula with an absolute value is clearly for real arguments. Because the complex function $\dfrac1x$ is analytical except at the origin, while $\log|x|$ is nowhere.
Hence the Problem 1 is irrelevant.
And for Problem 2, the formula without the absolute value would be of no use in the negatives if complex are disallowed.

This said, the formula $\log x$ would be better for the undefinite integral evaluated as 
$$\log\frac ba$$ instead of $$\log\left|\frac ba\right|$$ (and not as a difference of logarithms) as this forbids to straddle the origin.
A: So, two separate points: First, let me discuss your first example:
In the first series, you are forgetting that the sum you try to evaluate (for $z=e^{i\pi}\Rightarrow|z|=1$) lies on the border of the circle of convergence of the power series $p(z)=\sum_{n=1}^\infty\frac{z^n}n$ of $f(z)=-\ln(1-z).$ Therefore, we cannot say $$\sum_{n=1}^\infty\frac{e^{in\pi}}n\overset{!}=-\ln|1-e^{i\pi}|,$$ since convergence is not guaranteed for values on the border of the unit circle $\mathbb S^1.$
Secondly, you are correct: The formula $\int_a^b\frac1x\text dx=\ln|b|-\ln|a|$ is true for $0<a<b\vee a<b<0$, not generally. Once you define the logarithm of complex values, you should interpret the formula $\int_a^b\frac1x\text dx=\ln|x|$ as a "special case formula" for real numbers (such as $a^2+b^2=c^2$, which is only true for right-angled triangles, but has the more general cosine-formula for all triangles in the Euclidean plane) that allows you to compute any proper Riemann integral of the inverse function as long as the interval you are trying to integrate does not contain a neighbourhood of zero.
The advantage with this definition using $\ln\circ\mathop{\mathrm{abs}}$ is that you do not have to introduce any complex functions or the $\arg$ function, much less get bogged down in the principal value problem just to define the integral of an odd function on the negative part of its domain. After all, for $a<b<0,$ the equality holds. For $0<a<b$ trivially as well. However, unlike many other equalities that hold on most of $\mathbb R,$ this one is not extendable to any set $M\subset\mathbb C\setminus\mathbb R.$
Another example for such a formula would be $\forall a,b>0\colon\sqrt{ab}=\sqrt a\sqrt b.$
