#### Answer

$$\ln \left( {\frac{{16}}{9}} \right)$$

#### Work Step by Step

$$\eqalign{
& \int_{\ln 2}^{\ln 3} {\coth s} ds \cr
& {\text{hyperbolic function}} \cr
& = \int_{\ln 2}^{\ln 3} {\frac{{\cosh s}}{{\sinh s}}} ds \cr
& {\text{substitute }}u = \sinh s,{\text{ }}du = \cosh sds \cr
& {\text{express the limits in terms of }}u \cr
& x = \ln 3{\text{ implies }}u = \sinh s = 4/3 \cr
& x = \ln 2{\text{ implies }}u = \sinh s = 3/4 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_{\ln 2}^{\ln 3} {\frac{{\cosh s}}{{\sinh s}}} ds = \int_{3/4}^{4/3} {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\ln \left| u \right|} \right)} \right|_{3/4}^{4/3} \cr
& = \ln \left( {\frac{4}{3}} \right) - \ln \left( {\frac{3}{4}} \right) \cr
& = \ln \left( {\frac{{16}}{9}} \right) \cr} $$