Whether derivative of $\ln(x)$ is $\frac{1}{x}$ for $x>0$ only? Whether derivative of $\ln(x)$ is $\frac{1}{x}$ for $x>0$ only? Can't we write $$\frac{d}{dx} {\ln|x|} = \frac{1}{x} $$ so that we can get the corresponding integration formula for $\frac{1}{x}$ easily as $$\ln|x|$$
I have gone thorough this but it discusses only about integration Is the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$? 

Since $|x|=\sqrt{x^2}$ we have $\bigl(\ln|x|\bigr)'=\bigr(\ln(\sqrt{x^2})\bigr)'=\dfrac{1}{\sqrt{x^2}}\dfrac{1}{2\sqrt{x^2}}\cdot2x=\dfrac{x}{|x|^2}=\dfrac{x}{x^2}=\dfrac{1}{x}$.
