Calculus (3rd Edition)

$-1$
Using the fact that $\frac{u'}{u}=\ln u$, we have $$\int_{-e^2}^{-e} \frac{1}{t}dt= \ \ln( |t|)|_{-e^2}^{-e} \\ =\ln e-\ln e^2\\ =\ln e-2\ln e=1-2=-1.$$ where used the properties $\ln e=1$ and $\ln B^x=x\ln B$.