Answer
$$L \approx 6.7886512$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \ln x{\text{ on the interval }}\left[ {1,{e^2}} \right] \cr
& {\text{Therefore,}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{x} \cr
& {\text{The Arc length is given by:}} \cr
& L = \int_1^{{e^2}} {\sqrt {1 + {{\left( {\frac{1}{x}} \right)}^2}} dx} \cr
& L = \int_1^{{e^2}} {\frac{{\sqrt {{x^2} + 1} }}{x}dx} \cr
& {\text{Integrate by using the formula }} \cr
& \int {\frac{{\sqrt {{x^2} + {a^2}} }}{x}dx} = \sqrt {{x^2} + {a^2}} - a\ln \left| {\frac{{a + \sqrt {{x^2} + {a^2}} }}{x}} \right| + C \cr
& L = \left[ {\sqrt {{x^2} + 1} - \ln \left| {\frac{{1 + \sqrt {{x^2} + 1} }}{x}} \right|} \right]_1^{{e^2}} \cr
& L = \left[ {\sqrt {{{\left( {{e^2}} \right)}^2} + 1} - \ln \left| {\frac{{1 + \sqrt {{{\left( {{e^2}} \right)}^2} + 1} }}{{{e^2}}}} \right|} \right] - \left[ {\sqrt 2 - \ln \left| {\frac{{1 + \sqrt 2 }}{1}} \right|} \right] \cr
& {\text{Simplifying}} \cr
& L = \left[ {\sqrt {{e^4} + 1} - \ln \left| {\frac{{1 + \sqrt {{e^4} + 1} }}{{{e^2}}}} \right|} \right] - \left[ {\sqrt 2 - \ln \left| {1 + \sqrt 2 } \right|} \right] \cr
& L = \sqrt {{e^4} + 1} - \ln \left| {\frac{{1 + \sqrt {{e^4} + 1} }}{{{e^2}}}} \right| - \sqrt 2 + \ln \left| {1 + \sqrt 2 } \right| \cr
& L = \sqrt {{e^4} + 1} + \ln \left| {\frac{{{e^2}\left( {1 + \sqrt 2 } \right)}}{{1 + \sqrt {{e^4} + 1} }}} \right| - \sqrt 2 \cr
& L \approx 6.7886512 \cr} $$
