Answer
$$A = 6\ln \left( {\frac{{5 + \sqrt {21} }}{{3 + \sqrt 5 }}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{6}{{\sqrt {{x^2} - 4} }} \cr
& {\text{The area of the region is given by }} \cr
& A = \int_a^b {\left( {f\left( x \right) - g\left( x \right)} \right)} dx,{\text{ }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right],{\text{ then}} \cr
& A = \int_3^5 {\frac{6}{{\sqrt {{x^2} - 4} }}} dx \cr
& A = 6\int_3^5 {\frac{1}{{\sqrt {{x^2} - 4} }}} dx \cr
& {\text{Using theorem 5}}{\text{.20 }}\int {\frac{{du}}{{\sqrt {{u^2} - {a^2}} }} = \ln \left( {u + \sqrt {{u^2} - {a^2}} } \right) + C} \cr
& A = 6\left[ {\ln \left( {x + \sqrt {{x^2} - 4} } \right)} \right]_3^5 \cr
& A = 6\left[ {\ln \left( {5 + \sqrt {{{\left( 5 \right)}^2} - 4} } \right) - \ln \left( {3 + \sqrt {{{\left( 3 \right)}^2} - 4} } \right)} \right] \cr
& A = 6\left[ {\ln \left( {5 + \sqrt {21} } \right) - \ln \left( {3 + \sqrt 5 } \right)} \right] \cr
& {\text{By logarithmic properties}} \cr
& A = 6\ln \left( {\frac{{5 + \sqrt {21} }}{{3 + \sqrt 5 }}} \right) \cr} $$
