The antiderivative of 1/x is ln(x), not ln |x|

The antiderivative of $\frac{1}{x}$ is commonly shown to be $\ln \left|x\right|$, since the graph of $\ln x$ doesn't show anything where $x<0$. However, looking more closely, $\ln \left|x\right|$ can be rewritten as follows:

$\ln \left|x\right|=\{\begin{array}{ll}\ln (-x)& \text{if }x<0\\ \ln x& \text{if }x>0\end{array}$

Additionally, since an antiderivative can be offset by an arbitrary constant, $\ln \left|x\right|+3$ is also a valid antiderivative. But going back to the piecewise function — since $\frac{1}{x}$ is undefined at $x=0$, each piece can be offset by a different constant, like this:

$\int \frac{1}{x}dx=\{\begin{array}{ll}\ln (-x)+5& \text{if }x<0\\ \ln x+3& \text{if }x>0\end{array}$

Using this fact, I could choose the constants in a specific way:

$\int \frac{1}{x}dx=\{\begin{array}{ll}\ln (-x)+i\pi & \text{if }x<0\\ \ln x& \text{if }x>0\end{array}$

This is just as valid as an antiderivative, and is equivalent to taking the principal branch of $\ln x$ for negative values of $x$.

In fact, if the domain is extended to complex numbers, $\ln \left|x\right|$ is not differentiable at all in complex analysis, leaving only $\ln x$ as the antiderivative.