Answer
$$\frac{1}{4}\ln \left( {\frac{{29}}{{121}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{\ln 5}^{\ln 9} {\frac{{\cosh x}}{{4 - {{\sinh }^2}x}}} dx \cr
& {\text{set }}u = \sinh x{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = \cosh xdx, \cr
& \cr
& {\text{switch the limits of integration}} \cr
& u = \sinh x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \ln 5 \to u = 12/5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \ln 9 \to u = 40/9 \cr
& {\text{use the change of variable}} \cr
& \int_{\ln 5}^{\ln 9} {\frac{{\cosh x}}{{4 - {{\sinh }^2}x}}} dx = \int_{12/5}^{40/9} {\frac{{du}}{{{2^2} - {u^2}}}} \cr
& {\text{using the Integral formulas in the Theorem 6}}{\text{.12}}:{\text{ }}\int {\frac{{dx}}{{{a^2} - {x^2}}} = \frac{1}{a}{{\tanh }^{ - 1}}\frac{x}{a} + C,{\text{ for }}\left| x \right| < a} \cr
& or{\text{ }}\int {\frac{{dx}}{{{a^2} - {x^2}}}} = \frac{1}{a}{\coth ^{ - 1}}\frac{x}{a} + C,{\text{ for }}\left| x \right| > a \cr
& {\text{Comparing the integral}}\int_{12/5}^{40/9} {\frac{{du}}{{{2^2} - {u^2}}}} {\text{ with }}\int {\frac{{dx}}{{{a^2} - {x^2}}}} {\text{ we can note that }} \cr
& a = 2,{\text{ }}u = x{\text{ and the limits of the integration are }}\frac{{12}}{5} \leqslant u \leqslant \frac{{40}}{9}{\text{ then }}\left| u \right| > 2. \cr
& {\text{ So }}\int {\frac{{dx}}{{{a^2} - {x^2}}}} = \frac{1}{a}{\coth ^{ - 1}}\frac{x}{a} + C,{\text{ for }}\left| x \right| > a{\text{ then}} \cr
& \int_{12/5}^{40/9} {\frac{{du}}{{{2^2} - {u^2}}}} = \left( {\frac{1}{2}{{\coth }^{ - 1}}\frac{u}{2}} \right)_{12/5}^{40/9} \cr
& {\text{evaluate the limits}} \cr
& = \frac{1}{2}{\coth ^{ - 1}}\frac{{40/9}}{2} - \frac{1}{2}{\coth ^{ - 1}}\frac{{12/5}}{2} \cr
& = \frac{1}{2}{\coth ^{ - 1}}\frac{{20}}{9} - \frac{1}{2}{\coth ^{ - 1}}\frac{6}{5} \cr
& {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}:{\text{ }} \cr
& {\coth ^{ - 1}}x = {\tanh ^{ - 1}}\left( {\frac{1}{x}} \right){\text{ for }}\left( {\left| x \right| > 1} \right).{\text{ Then}}{\text{,}} \cr
& = \frac{1}{2}{\tanh ^{ - 1}}\frac{9}{{20}} - \frac{1}{2}{\tanh ^{ - 1}}\frac{5}{6} \cr
& {\text{And from the theorem 6}}{\text{.10 we have that }}{\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right){\text{ }}\left( {\left| x \right| < 1} \right){\text{ then}}{\text{,}} \cr
& = \frac{1}{2}\left( {\frac{1}{2}\ln \left( {\frac{{1 + 9/20}}{{1 - 9/20}}} \right)} \right) - \frac{1}{2}\left( {\frac{1}{2}\ln \left( {\frac{{1 + 5/6}}{{1 - 5/6}}} \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{4}\ln \left( {\frac{{29}}{{11}}} \right) - \frac{1}{4}\ln \left( {11} \right) \cr
& = \frac{1}{4}\ln \left( {\frac{{29/11}}{{11}}} \right) \cr
& = \frac{1}{4}\ln \left( {\frac{{29}}{{121}}} \right) \cr} $$
