Answer
$$L = \frac{{35}}{{24}} + 3\ln 6$$
Work Step by Step
$$\eqalign{
& y = 3\ln x - \frac{{{x^2}}}{{24}}{\text{ on }}\left[ {1,6} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = 3\ln x - \frac{{{x^2}}}{{24}}{\text{ and }}\left[ {1,6} \right] \to a = 1{\text{ and }}b = 6.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {3\ln x - \frac{{{x^2}}}{{24}}} \right] \cr
& f'\left( x \right) = 3\left( {\frac{1}{x}} \right) - \frac{{2x}}{{24}} \cr
& f'\left( x \right) = \frac{3}{x} - \frac{x}{{12}} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_1^6 {\sqrt {1 + {{\left( {\frac{3}{x} - \frac{x}{{12}}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_1^6 {\sqrt {1 + \frac{9}{{{x^2}}} - 2\left( {\frac{3}{x}} \right)\left( {\frac{x}{{12}}} \right) + {{\left( {\frac{x}{{12}}} \right)}^2}} } dx \cr
& L = \int_1^6 {\sqrt {1 + \frac{9}{{{x^2}}} - \frac{1}{2} + \frac{{{x^2}}}{{144}}} } dx \cr
& L = \int_1^6 {\sqrt {\frac{{{x^2}}}{{144}} + \frac{1}{2} + \frac{9}{{{x^2}}}} } dx \cr
& {\text{factoring }} \cr
& L = \int_1^6 {\sqrt {{{\left( {\frac{x}{{12}} + \frac{3}{x}} \right)}^2}} } dx \cr
& L = \int_1^6 {\left( {\frac{x}{{12}} + \frac{3}{x}} \right)} dx \cr
& {\text{integrate and evaluate}} \cr
& L = \left( {\frac{{{x^2}}}{{24}} + 3\ln \left| x \right|} \right)_1^6 \cr
& L = \left( {\frac{{{6^2}}}{{24}} + 3\ln \left| 6 \right|} \right) - \left( {\frac{{{1^2}}}{{24}} + 3\ln \left| 1 \right|} \right) \cr
& L = \frac{{36}}{{24}} + 3\ln 6 - \frac{1}{{24}} \cr
& L = \frac{{35}}{{24}} + 3\ln 6 \cr} $$
