Is there a proof that $\int \frac {dx}{x}=\ln |x|+c$? Is there a proof that $$\int \frac {dx}{x}= \ln|x|+c$$
for $x\neq 0$
I would be interest for any replies or any comment.
 A: If we accept that $\frac{d}{dx}e^x=e^x$, then we can simply use implicit differentiation.
$$\begin{align}y&=\ln x\\x&=e^y\\\frac{d}{dx}x&=\frac d {dx}e^y\\1&=\frac{dy}{dx}e^y\\\frac 1{e^y}&=\frac{dy}{dx}\\\frac{dy}{dx}&=\frac 1 x\end{align}$$
A: Suppose we want to find the antiderivative of $\cfrac 1x$ - it doesn't come in the pattern of derivatives of powers, because differentiating a constant always gives zero. When we look at the function we realise we will have a problem if our interval of integration includes zero, so we have to avoid that. And we see that this is an odd function, so if we have negative limits, we change them to positive and change the sign. So we deal with positive limits throughout and adapt to negative limits later.
Let's start thinking about the properties of $$\int_a^x \cfrac 1y dy=\int_a^1 \cfrac 1y dy+\int_1^x \cfrac 1y dy=c+\int_1^x \cfrac 1y dy$$ so we might as well have the lower limit $1$ and consider $$L(x)=\int_1^x \cfrac 1y dy$$
We note that $L(1)=0$ and $$L(ab)=\int_1^{ab} \cfrac 1x dx=\int_1^a \cfrac 1x dx+\int_a^{ab} \cfrac 1x dx=L(a)+L(b)$$ after a simple substitution in the second integral.
Then the properties of a logarithm are established and the base is the number $e$ such that $L(e)=1$. So maybe you are looking for a proof that this $e$ is the same one you get by some other route?
This was too long for a comment.
A: Is your question where the absolute value comes in?
for $x > 0$, $|x| = x$, in which case
$D_x \ln |x| = D_x \ln x = \dfrac 1 x$
for $x < 0$, then $|x| = -x$, and
$D_x \ln |x| = D_x \ln (-x) = -1 \cdot \dfrac 1 {-x} = \dfrac 1 x$
A: The logarithm can be defined by
$$\ln x=\lim_{n\to\infty}n(\sqrt[n]x-1).$$
Then deriving,
$$(\ln x)'=\lim_{n\to\infty}n\frac1nx^{1/n-1}=\frac1x.$$
(This is not a rigorous proof.)
A: In all fairness, the formula $\displaystyle\int x^{-1}dx=\log |x|+c$ is incorrect, or at least is unexhaustive. The correct formula is  $$=\begin{cases}\log x+c_1;x>0\\\log(-x)+c_2;x<0\end{cases}$$ where $c_1,c_2$ are arbitrary constants. 
The point is that the singular point at $x=0$ makes the solution space of $xy'=1$ two dimensional instead of $1$ dimensional, as one would expect in a first degree equation. 
