For some time, I've seen different solutions for the same problems.
Let $f$ be any continuous function, differenciable on its domain such that,
$$\int \frac{\mathrm d f}{f}=\ln(f)$$
but some authors say,
$$\int \frac{\mathrm d f}{f}=\ln|f|$$
I know that,
$$\displaystyle \frac{\mathrm d}{\mathrm dx}\ln |x|=\frac{\mathrm d}{\mathrm dx}\ln \sqrt(x^2)=$$ $$\displaystyle \frac{\mathrm d}{\mathrm dx}\ln (x^2)^{\frac{1}{2}}=\frac{\mathrm d}{\mathrm dx}\frac{1}{2}\ln (x^2)=$$ $$\frac{\mathrm d}{\mathrm dx}\ln (x)$$
And so I concluded that: $$\frac{\mathrm d}{\mathrm dx} \ln|x|=\frac{\mathrm d}{\mathrm dx} \ln(x)$$
Can I generalizate that,
$$\int \frac{\mathrm d f}{f}=\ln(f)= \ln|f|$$
The functions $\ln x$ and $\ln |x|$ are defined on different regions: The former is defined for positive $x$ while the latter is defined for every non-zero $x$.
Using the chain rule, and the fact the derivative of the absolute value is the signum function one can see that $$\frac{d}{dx} \ln |f(x)| =\frac{1}{|f(x)|} (\text{sgn} f(x)) f'(x).$$ Since $\text{sgn}=\pm1$ we can move it to the denominator, so $$\frac{d}{dx} \ln |f(x)| =\frac{1}{(\text{sgn} f(x))|f(x)|} f'(x).$$
And lastly, for any real $t$, $|t| \cdot \text{sgn}t=t$. Thus $$\frac{d}{dx} \ln |f(x)| =\frac{f'(x)}{f(x)}.$$
Which coincides with the derivative of $\ln f(x)$.
To summarize: $\ln f$ and $\ln |f|$ have the same derivative but the second one is defined over a greater set.