## Calculus: Early Transcendentals (2nd Edition)

$$2\ln \left| x \right| - 3\ln \left| {x + 1} \right| + \ln \left| {x - 1} \right| + C$$
\eqalign{ & \int {\frac{{4x - 2}}{{{x^3} - x}}dx} \cr & {\text{integrand }} \cr & = \frac{{4x - 2}}{{{x^3} - x}} \cr & {\text{factor}} \cr & = \frac{{4x - 2}}{{x\left( {{x^2} - 1} \right)}} \cr & = \frac{{4x - 2}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} \cr & {\text{partial fractions}} \cr & \frac{{4x - 2}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 1}} + \frac{C}{{x - 1}} \cr & 4x - 2 = A\left( {x + 1} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right) + Cx\left( {x + 1} \right) \cr & {\text{Letting }}x = 0 \cr & 4\left( 0 \right) - 2 = A\left( {0 + 1} \right)\left( {0 - 1} \right) + B\left( 0 \right) + C\left( 0 \right) \cr & - 2 = A\left( { - 1} \right) \cr & A = 2 \cr & {\text{Letting }}x = - 1 \cr & 4\left( { - 1} \right) - 2 = A\left( 0 \right) + B\left( { - 1} \right)\left( { - 1 - 1} \right) + C\left( 0 \right) \cr & - 6 = B\left( 2 \right) \cr & B = - 3 \cr & {\text{Letting }}x = 1 \cr & 4\left( 1 \right) - 2 = A\left( 0 \right) + B\left( 0 \right) + C\left( 1 \right)\left( {1 + 1} \right) \cr & 2 = C\left( 2 \right) \cr & C = 1 \cr & {\text{substituting constants}} \cr & \frac{A}{x} + \frac{B}{{x + 1}} + \frac{C}{{x - 1}} = \frac{2}{x} + \frac{{ - 3}}{{x + 1}} + \frac{1}{{x - 1}} \cr & \int {\frac{{4x - 2}}{{{x^3} - x}}dx} = \int {\left( {\frac{2}{x} + \frac{{ - 3}}{{x + 1}} + \frac{1}{{x - 1}}} \right)dx} \cr & {\text{integrating}} \cr & = 2\ln \left| x \right| - 3\ln \left| {x + 1} \right| + \ln \left| {x - 1} \right| + C \cr}