$$\int_{20}^{10} -1/x dx$$
$ \begin{align*}\int_{20}^{10} -1/x dx&=\int_{10}^{20} 1/x dx \\&=\left. {\log(x)} \right|_{10}^{20}\\&=\log(2) \end{align*}$
It was by no means a fluke. Anyone experienced enough would see that the domain of integration is a subset of the positive real numbers, therefore the absolute values is not necessary.
Moreover, when doing an integral as apposed to an antederivative one need not use a C.
You don't know what you're talking about. The *correct* antiderivative formula is as I stated with the absolute value bars. That is a separate issue from continuing into the next part
of the problem. Anyone experienced enough knows that. Therefore, *after* noticing that, then the qualification may be stated that parentheses are sufficient where it is placed into
the definite integral. So, yes, it was by means a fluke, because the argument without the absolute value didn't change anything. Please be careful next time with your logic.
Moreover, in my haste to correct him, I left off the "+ C" for the indefinite integral. It should be:
$\displaystyle \int -\dfrac 1 x~dx = -\ln|x| + C$