Answer
$$\kappa \left( x \right) = \frac{x}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln x \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}f''\left( x \right) \cr
& f'\left( x \right) = \frac{1}{x} \cr
& f''\left( x \right) = - \frac{1}{{{x^2}}} \cr
& {\text{Use the formula }}\kappa \left( x \right) = \frac{{\left| {f''\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \left| { - \frac{1}{{{x^2}}}} \right|\frac{1}{{{{\left( {1 + {{\left( {1/x} \right)}^2}} \right)}^{3/2}}}} \cr
& {\text{Simplifying}} \cr
& \kappa \left( x \right) = \left( {\frac{1}{{{x^2}}}} \right)\frac{1}{{{{\left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \left( {\frac{1}{{{x^2}}}} \right)\frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \frac{x}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr} $$
