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