Answer
$$L = \int_1^{10} {\frac{1}{x}\sqrt {{x^2} + 1} } dx$$
Work Step by Step
$$\eqalign{
& y = \ln x{\text{ on }}\left[ {1,10} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = \ln x{\text{ and }}\left[ {1,10} \right] \to a = 1{\text{ and }}b = 10.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x} \right] \cr
& f'\left( x \right) = \frac{1}{x} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_1^{10} {\sqrt {1 + {{\left( {\frac{1}{x}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_1^{10} {\sqrt {1 + \frac{1}{{{x^2}}}} } dx \cr
& L = \int_1^{10} {\sqrt {\frac{{{x^2} + 1}}{{{x^2}}}} } dx \cr
& L = \int_1^{10} {\frac{1}{x}\sqrt {{x^2} + 1} } dx \cr} $$
