Answer
$$L = 3\sqrt 2 - \sqrt {10} + 3\ln \left( {\frac{{3 + \sqrt {10} }}{{1 + \sqrt 2 }}} \right)$$
Work Step by Step
$$\eqalign{
& y = 3\ln x,{\text{ 1}} \leqslant x \leqslant 3 \cr
& {\text{Calculate the arc length of the curve using the formula}}{\text{.}} \cr
& L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& {\text{,then}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {3\ln x} \right] = \frac{3}{x} \cr
& L = \int_1^3 {\sqrt {1 + {{\left( {\frac{3}{x}} \right)}^2}} dx} \cr
& L = \int_1^3 {\sqrt {1 + \frac{9}{{{x^2}}}} dx} \cr
& L = \int_1^3 {\frac{1}{x}\sqrt {{x^2} + 9} dx} \cr
& {\text{Integrate using the formula }} \cr
& \int {\frac{{\sqrt {{u^2} + {a^2}} }}{u}du = \sqrt {{u^2} + {a^2}} - a\ln \left| {\frac{{a + \sqrt {{u^2} + {a^2}} }}{u}} \right| + C} \cr
& L = \left[ {\sqrt {{x^2} + 9} - 3\ln \left| {\frac{{3 + \sqrt {{x^2} + 9} }}{x}} \right|} \right]_1^3 \cr
& L = \left[ {\sqrt {{3^2} + 9} - 3\ln \left| {\frac{{3 + \sqrt {{3^2} + 9} }}{3}} \right|} \right] - \left[ {\sqrt {10} - \ln \left| {\frac{{3 + \sqrt {10} }}{3}} \right|} \right] \cr
& {\text{Simplifying}} \cr
& L = \left[ {\sqrt {18} - 3\ln \left| {\frac{{3 + \sqrt {18} }}{3}} \right|} \right] - \left[ {\sqrt {10} - \ln \left| {\frac{{3 + \sqrt {10} }}{3}} \right|} \right] \cr
& L = 3\sqrt 2 - 3\ln \left| {\frac{{3 + 3\sqrt 2 }}{3}} \right| - \sqrt {10} + \ln \left| {\frac{{3 + \sqrt {10} }}{3}} \right| \cr
& L = 3\sqrt 2 - 3\ln \left( {1 + \sqrt 2 } \right) - \sqrt {10} + \ln \left( {\frac{{3 + \sqrt {10} }}{3}} \right) \cr
& {\text{Using logarithmic properties}} \cr
& L = 3\sqrt 2 - \sqrt {10} + 3\ln \left( {\frac{{3 + \sqrt {10} }}{{1 + \sqrt 2 }}} \right) \cr} $$
