Answer
$$3x + \frac{1}{2}\ln \left| {{x^2} - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^2} + x - 3}}{{{x^2} - 1}}} \cr
& {\text{By the long division }}\frac{{3{x^2} + x - 3}}{{{x^2} - 1}} = 3 + \frac{x}{{{x^2} - 1}} \cr
& {\text{Therefore,}} \cr
& \int {\frac{{3{x^2} + x - 3}}{{{x^2} - 1}}} dx = \int {\left( {3 + \frac{x}{{{x^2} - 1}}} \right)dx} \cr
& = \int {3dx} + \frac{1}{2}\int {\frac{{2x}}{{{x^2} - 1}}dx} \cr
& {\text{Integrate }} \cr
& {\text{ = }}3x + \frac{1}{2}\ln \left| {{x^2} - 1} \right| + C \cr} $$
