Answer
$$\frac{3}{2}\ln \left| {{x^2} - 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ - x}}{{2 - {x^2}}}} dx \cr
& {\text{rewrite the integrand}} \cr
& \int {\frac{x}{{{x^2} - 2}}} dx \cr
& {\text{set }}u = {x^2} - 2{\text{ then }}\frac{{du}}{{dx}} = 2x,\,\,\,\,\,\,\,\,xdx = \frac{1}{2}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{x}{{{x^2} - 2}}} dx = \int {\frac{{1/2}}{u}} du \cr
& = \frac{1}{2}\int {\frac{1}{u}} du \cr
& {\text{integrate by using }}\int {\frac{1}{u}} du = \ln \left| u \right| + C \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{replace }}{x^2} - 2{\text{ for }}u \cr
& = \frac{3}{2}\ln \left| {{x^2} - 2} \right| + C \cr} $$