text: Q:
  
  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. 
  
  