## Calculus: Early Transcendentals (2nd Edition)

$$2$$
\eqalign{ & \int_1^{{e^2}} {\frac{{\ln p}}{p}dp} \cr & {\text{substitute }}u = \ln p,{\text{ }}du = \frac{1}{p}dp \cr & {\text{express the limits in terms of }}u \cr & x = 1{\text{ implies }}u = \ln \left( 1 \right) = 0 \cr & x = {e^2}{\text{ implies }}u = \ln \left( {{e^2}} \right) = 2 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_1^{{e^2}} {\frac{{\ln p}}{p}dp} = \int_0^2 {udu} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\frac{{{u^2}}}{2}} \right)} \right|_0^2 \cr & {\text{use the fundamental theorem}} \cr & = \frac{{{2^2}}}{2} - \frac{{{0^2}}}{2} \cr & {\text{simplify}} \cr & = 2 \cr}