Answer
$${\text{2ln}}\left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{4}{{x + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{8x}}{{{x^3} + {x^2} - x - 1}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{8x}}{{{x^3} + {x^2} - x - 1}} = \frac{{8x}}{{{x^2}\left( {x + 1} \right) - \left( {x + 1} \right)}} = \frac{{8x}}{{\left( {{x^2} - 1} \right)\left( {x + 1} \right)}} \cr
& = \frac{{8x}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{{{\left( {x + 1} \right)}^2}}} \cr
& 8x = A{\left( {x + 1} \right)^2} + B\left( {x + 1} \right)\left( {x - 1} \right) + C\left( {x - 1} \right) \cr
& 8x = A{x^2} + 2Ax + A + B{x^2} - B + Cx - C \cr
& 8x = \left( {A + B} \right){x^2} + \left( {2A + C} \right)x + \left( {A - B - C} \right) \cr
& A + B = 0 \cr
& 2A + C = 8 \cr
& A - B - C = 0 \cr
& {\text{Solving the system of equations}} \cr
& A = 2,{\text{ }}B = - 2,C{\text{ = 4}} \cr
& \frac{{8x}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}} = \frac{2}{{x - 1}} + \frac{{ - 2}}{{x + 1}} + \frac{4}{{{{\left( {x + 1} \right)}^2}}} \cr
& \int {\frac{{8x}}{{{x^3} + {x^2} - x - 1}}} dx = \int {\left( {\frac{2}{{x - 1}} + \frac{{ - 2}}{{x + 1}} + \frac{4}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx \cr
& {\text{Integrate}} \cr
& {\text{ = 2ln}}\left| {x - 1} \right| - 2\ln \left| {x + 1} \right| - \frac{4}{{x + 1}} + C \cr
& {\text{ = 2ln}}\left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{4}{{x + 1}} + C \cr} $$
