This example comes from the Z-transform chapter of "Signals and Linear Systems" (3rd edition) by Gabel & Roberts. I can't quite replicate their result. The book states
$$-\int_0^z \frac{z^{-2}}{1-{z^{-1}}}\;\mathrm{d\hat{z}}=-\ln(1-z^{-1})$$ I believe this may be re-written as $$-\int_0^z \frac{1}{z-{z^2}}\;\mathrm{d\hat{z}}$$
Note that we have the condition $1<|z|$
Using hints from Wolfram (without Go Pro), I then re-phrased the problem as $$-\int_0^z \frac{1}{\left(\frac{1}{2}\right)^2-\left(z-\frac{1}{2}\right)^2}\;\mathrm{d\hat{z}}$$
(The above should be better formatted, not sure how to get the correct relative 'heights' yet)
Then using a table of integrals, e.g. here (https://math24.net/integration-completing-square.html) $$-\int \frac{1}{a^2-z^2}\;\mathrm{dx}=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|$$ and writing $u=z-\frac{1}{2},du=d\hat{z},a=\frac{1}{2}$ $$-\int_0^z \frac{d\hat{z}}{(\frac{1}{2})^2-u^2}\;=\ln\left| \frac{\frac{1}{2}+(z-\frac{1}{2})}{\frac{1}{2}-\left(z-\frac{1}{2}\right)} \right|$$ $$=\ln\left|\frac{1}{z}-1\right|$$ as the indefinite integral
This has the opposite sign to the book's stated result, and I do not know if this is because the book is showing the results once the limits are substituted.
However, I am unclear about using the given limits. Is the lower limit that shown for the integral i.e. 0, or maybe "just a little bit over 1" due to the convergence range given for z?
Thanks for the title edit, and the comments. Strictly, the book uses $$\int_0^zf(z)d\hat{z}$$ with the comment "We note that $\hat{z}$ in this expression may be treated as if it were a real variable".
I've now changed the integrals to refect this. Maybe one error I made was to assume that I could write "$dz$" in place of "$d\hat{z}$" in light of that remark.