Answer
$$\frac{{37}}{{375}}$$
Work Step by Step
$$\eqalign{
& \int_{\ln 2}^{\ln 3} {\tanh x{{\operatorname{sech} }^3}x} dx \cr
& {\text{split}} \cr
& \int_{\ln 2}^{\ln 3} {sec{h^2}x\tanh x\operatorname{sech} x} dx \cr
& {\text{substitute }}u = \operatorname{sech} x,{\text{ }}du = - \operatorname{sech} x\tanh xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \ln 2,{\text{ }}u = \operatorname{sech} \left( {\ln 2} \right) = 4/5 \cr
& {\text{if }}x = \ln 3,{\text{ }}u = \operatorname{sech} \left( {\ln 3} \right) = 3/5 \cr
& {\text{so}} \cr
& \int_{\ln 2}^{\ln 3} {sec{h^2}x\tanh x\operatorname{sech} x} dx = - \int_{4/5}^{3/5} {{u^2}} du \cr
& {\text{find the antiderivative}} \cr
& = - \left. {\left( {\frac{{{u^3}}}{3}} \right)} \right|_{4/5}^{3/5} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \left( {\frac{{{{\left( {3/5} \right)}^3}}}{3} - \frac{{{{\left( {4/5} \right)}^3}}}{3}} \right) \cr
& = \frac{{37}}{{375}} \cr} $$
