#### Answer

$$\ln 4$$

#### Work Step by Step

$$\eqalign{
& \int_{{e^2}}^{{e^8}} {\frac{{dx}}{{x\ln x}}} \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& {\text{express the limits in terms of }}u \cr
& x = {e^2}{\text{ implies }}u = \ln {e^2} = 2 \cr
& x = {e^8}{\text{ implies }}u = \ln {e^8} = 8 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_{{e^2}}^{{e^8}} {\frac{{dx}}{{x\ln x}}} = \int_2^8 {\frac{1}{u}} du \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\ln \left| u \right|} \right)} \right|_2^8 \cr
& {\text{use the fundamental theorem}} \cr
& = \ln \left| 8 \right| - \ln \left| 2 \right| \cr
& {\text{properties of logarithms}} \cr
& = \ln \left( {\frac{8}{2}} \right) \cr
& = \ln 4 \cr} $$