What is the proof that $$\int \frac{1}{x}\, \text{d}x=\ln |x|?$$
Please note that similar questions refer to $\ln x$, and not $\ln |x|$ as I have seen in Calculus textbooks.
Given the formula that $$\frac{\text{d}}{\text{d}x}\ln x=\frac{1}{x},$$ it is obvious that for $x>0$, $\int 1/x\, \text{d}x=\ln x$; however I am unsure how the above definition is proven over all $x\neq 0$.
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When $x\lt 0$, $\ln |x|=\ln(-x)$, but then:
$$\frac{d}{dx}\ln(-x)=\frac{1}{-x}\frac{d}{dx}(-x)=\frac{1}{-x}(-1)=\frac{1}{x}$$
so $\ln|x|$ is the antiderivative of $\frac{1}{x}$ for $x\lt 0$ too.
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I believe the easiest solution is to show that $\frac{d}{dx} \ln \lvert x \rvert = \frac{1}{x}$, and since you know that $\ln(x) = \ln \lvert x \rvert$ for $x >0$ you can treat this as an extension of the "normal" natural logarithm.
You can do this by using that $\lvert x \rvert = \sqrt{x^2}$ and using the chain rule. You get \begin{align*} \frac{d}{dx} \ln \lvert x \rvert & = \frac{d}{dx} \ln \left( \sqrt{x^2}\right)\\ & = \frac{1}{ \sqrt{x^2}}\cdot \frac{d}{dx} \sqrt{x^2}\\ & = \frac{1}{ \sqrt{x^2}}\cdot \frac{1}{2\sqrt{x^2}}\cdot\frac{d}{dx} x^2\\ & = \frac{1}{ \color{purple}{\sqrt{x^2}}}\cdot \frac{1}{\color{blue}{2}\color{purple}{\sqrt{x^2}}} \cdot \color{blue}{2}x\\ & = \frac{1}{\color{purple}{x^2}} \cdot x\\ &= \frac{1}{x} \end{align*}
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$$y=\ln|x|$$ $$e^y=|x|$$ $$e^y\frac{dy}{dx}=\operatorname{sgn}(x)$$ $$\frac{dy}{dx}=\frac{\operatorname{sgn}(x)}{|x|}=\frac1x$$ where the exponential function: $$e^y=\exp(y)$$ is defined as the function satisfying the following statements: $$f(x)=\exp(x),\,f^{-1}(x)=\ln(x)$$ $$f'=f$$
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An "indefinite integral" formula of the form $$ \int f(x)\,dx=F(x)+C\tag{1} $$ is understood as $F$ is an antiderivative on some (open) interval $I$, i.e., $F'(x)=f(x)$ for every $x\in I$. When the context is clear, one usually write (1) without mentioning explicitly the interval $I$.
The formula $$ \int \frac{1}{x}\,dx=\ln|x|+C $$ is understood valid for any (open) interval $I$ not containing $0$.
The natural logarithm function $\ln x$ is defined only for $x>0$. It does not make sense to prove $$ \frac{d}{dx}\ln(x)=\frac{1}{x},\quad x<0 $$ since the left hand side is undefined.
However, one can prove that for $x<0$,
$$ \frac{d}{dx}\ln(|x|)=\frac{1}{x}\tag{2} $$
You can prove (2) using the chain rule if you allow to use the fact $\frac{d}{dx}\ln(x)=\frac{1}{x}$ for $x>0$.
If you want to prove (2) from first principles, then you need to give the definition for the logarithm.
On $(0,\infty)$, we have $\log|x|=\log x$ and, as you know, $\log'(x)=\frac1x$.
And, on $(-\infty,0)$, $\log|x|=\log(-x)$ and, by the chain rule, if you differentiate $\log(-x)$, you get $-\frac1{-x}$, which is equal to $\frac1x$.