#### Answer

$$\frac{{32}}{3}$$

#### Work Step by Step

$$\eqalign{
& \int_1^{{e^2}} {\frac{{{{\left( {\ln x} \right)}^5}}}{x}dx} \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& {\text{express the limits in terms of }}u \cr
& x = {e^2}{\text{ implies }}u = \ln \left( {{e^2}} \right) = 2 \cr
& x = 1{\text{ implies }}u = \ln \left( 1 \right) = 0 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_1^{{e^2}} {\frac{{{{\left( {\ln x} \right)}^5}}}{x}dx} = \int_0^2 {{u^5}du} \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\frac{{{u^6}}}{6}} \right)} \right|_0^2 \cr
& {\text{evaluate limits}} \cr
& = \frac{{{2^6}}}{6} - \frac{{{0^6}}}{6} \cr
& {\text{simplify}} \cr
& = \frac{{64}}{6} \cr
& = \frac{{32}}{3} \cr} $$