Answer
$$7 + \ln \left( 8 \right)$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 2 }^3 {\frac{{2{x^3}}}{{{x^2} - 1}}} dx \cr
& {\text{By using long division }}\frac{{2{x^3}}}{{{x^2} - 1}} = 2x + \frac{{2x}}{{{x^2} - 1}} \cr
& \int_{\sqrt 2 }^3 {\frac{{2{x^3}}}{{{x^2} - 1}}} dx = \int_{\sqrt 2 }^3 {\left( {2x + \frac{{2x}}{{{x^2} - 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left( {{x^2} - \ln \left| {{x^2} - 1} \right|} \right)_{\sqrt 2 }^3 \cr
& = \left( {{{\left( 3 \right)}^2} + \ln \left| {{{\left( 3 \right)}^2} - 1} \right|} \right) - \left( {{{\left( {\sqrt 2 } \right)}^2} + \ln \left| {{{\left( {\sqrt 2 } \right)}^2} - 1} \right|} \right) \cr
& {\text{Simplifying, we get:}} \cr
& = \left( {9 + \ln \left( 8 \right)} \right) - \left( {2 + \ln \left( 1 \right)} \right) \cr
& = 9 + \ln \left( 8 \right) - 2 \cr
& = 7 + \ln \left( 8 \right) \cr} $$
