Answer
$$ - \ln \left| x \right| + \ln \left( {{x^2} + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} - 1}}{{{x^3} + x}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{{x^2} - 1}}{{{x^3} + x}} = \frac{{{x^2} - 1}}{{x\left( {{x^2} + 1} \right)}} \cr
& \frac{{{x^2} - 1}}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& {x^2} - 1 = A\left( {{x^2} + 1} \right) + B{x^2} + Cx \cr
& {x^2} - 1 = A{x^2} + A + B{x^2} + Cx \cr
& A = - 1 \cr
& Cx = 0 \to C = 0 \cr
& A + B = 1 \to B = 2 \cr
& \frac{{{x^2} - 1}}{{x\left( {{x^2} + 1} \right)}} = \frac{{ - 1}}{x} + \frac{{2x}}{{{x^2} + 1}} \cr
& \int {\frac{{{x^2} - 1}}{{{x^3} + x}}} dx = \int {\left( {\frac{{ - 1}}{x} + \frac{{2x}}{{{x^2} + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }} - \ln \left| x \right| + \ln \left( {{x^2} + 1} \right) + C \cr} $$
