## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{{\ln 2}} - \frac{1}{{\ln 3}}$$
\eqalign{ & \int_{{e^2}}^{{e^3}} {\frac{{dx}}{{x\ln x{{\ln }^2}\left( {\ln x} \right)}}} \cr & {\text{substitute }}u = \ln \left( {\ln x} \right),{\text{ }}du = \frac{{1/x}}{{\ln x}}dx \cr & {\text{express the limits in terms of }}u \cr & x = {e^3}{\text{ implies }}u = \ln \left( {\ln {e^3}} \right) = \ln 3 \cr & x = {e^2}{\text{ implies }}u = \ln \left( {\ln {e^2}} \right) = \ln 2 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{{e^2}}^{{e^3}} {\frac{{dx}}{{x\ln x{{\ln }^2}\left( {\ln x} \right)}}} = - \int_{\ln 2}^{\ln 3} {\frac{{du}}{{{u^2}}}} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right)} \right|_{\ln 2}^{\ln 3} \cr & = - \left. {\left( {\frac{1}{u}} \right)} \right|_{\ln 2}^{\ln 3} \cr & {\text{use the fundamental theorem}} \cr & = - \left( {\frac{1}{{\ln 3}} - \frac{1}{{\ln 2}}} \right) \cr & {\text{simplify}} \cr & = \frac{1}{{\ln 2}} - \frac{1}{{\ln 3}} \cr}