Answer
$$\frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{{x^2} + 1}}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{\sqrt {8 - 2{x^2} - {x^4}} }}} \cr
& {\text{Complete the square in the denominator}} \cr
& 8 - 2{x^2} - {x^4} = 9 - 1 - 2{x^2} - {x^4} \cr
& 8 - 2{x^2} - {x^4} = 9 - \left( {{x^4} + 2{x^2} + 1} \right) \cr
& 8 - 2{x^2} - {x^4} = 9 - {\left( {{x^2} + 1} \right)^2} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{xdx}}{{\sqrt {8 - 2{x^2} - {x^4}} }}} = \int {\frac{{xdx}}{{\sqrt {9 - {{\left( {{x^2} + 1} \right)}^2}} }}} \cr
& {\text{Let }}u = {x^2} + 1,\,\,\,du = 2xdx \cr
& \int {\frac{{xdx}}{{\sqrt {9 - {{\left( {{x^2} + 1} \right)}^2}} }}} = \int {\frac{{\left( {1/2} \right)du}}{{\sqrt {9 - {u^2}} }}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{\sqrt {{{\left( 3 \right)}^2} - {u^2}} }}} \cr
& \cr
& {\text{Integrating by tables, }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \frac{1}{2}\int {\frac{{du}}{{\sqrt {{{\left( 3 \right)}^2} - {u^2}} }}} = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr
& {\text{Write in terms of }}x{\text{; replace }}{x^2} + 1{\text{ for }}u \cr
& = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{{x^2} + 1}}{3}} \right) + C \cr} $$
