Answer
\[ = 2{e^3}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \int_e^x {\sqrt {{{\ln }^2}t - 1} \,\,dt} \hfill \\
\hfill \\
{\text{use }}ds = \int {\sqrt {1 + \,{{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \hfill \\
\hfill \\
\int_{}^{} {ds} = s = \int_e^{{e^3}} {\sqrt {1 + \,{{\left( {\frac{{df\left( x \right)}}{{dx}}} \right)}^2}} dx} \hfill \\
\hfill \\
then \hfill \\
\frac{{df\,\left( x \right)}}{{dx}} = \sqrt {{{\ln }^2}x - 1} \hfill \\
\hfill \\
s = \int_e^{{e^3}} {\sqrt {1 + {{\ln }^2}x - 1} } \,dx \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \int_e^{{e^3}} {\ln xdx} \hfill \\
\hfill \\
integrating\,\,by\,parts\,\, \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
u = \ln x \hfill \\
v = 1 \hfill \\
\hfill \\
S = \left. {x\ln x} \right|_e^{{e^3}} - \int_e^{{e^3}} {x \cdot \frac{1}{x}dx} \hfill \\
\hfill \\
= \,\left( {3{e^3} - e} \right) - \left. x \right|_e^{{e^3}} \hfill \\
\hfill \\
evaluate\,\,the\,\,{\text{limits}} \hfill \\
\hfill \\
= 3{e^3} - e - {e^3} + e = 2{e^3} \hfill \\
\hfill \\
\end{gathered} \]
