## Calculus: Early Transcendentals (2nd Edition)

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