Chapter 7 - Principles Of Integral Evaluation - 7.5 Integrating Rational Functions By Partial Fractions - Exercises Set 7.5 - Page 521: 14

$$ - \ln \left| x \right| + \frac{1}{2}\ln \left| {\frac{{x + 1}}{{x - 1}}} \right| + C$$

$$\eqalign{ & \int {\frac{{dx}}{{x\left( {{x^2} - 1} \right)}}} \cr & {\text{Decomposing the integrand into partial fractions}} \cr & {\text{Factor the denominator}} \cr & \frac{1}{{x\left( {{x^2} - 1} \right)}} = \frac{1}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} \cr & {\text{The form of the partial fraction decomposition is}} \cr & \frac{1}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 1}} + \frac{C}{{x - 1}} \cr & {\text{Multiplying the equation by }}x\left( {x + 1} \right)\left( {x - 1} \right){\text{, we have}} \cr & 1 = A\left( {x + 1} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right) + Cx\left( {x + 1} \right) \cr & {\text{if we set }}x = 0 \cr & 1 = A\left( 1 \right)\left( { - 1} \right) + B\left( 0 \right) + C\left( 0 \right) \cr & A = - 1 \cr & {\text{if we set }}x = - 1 \cr & 1 = A\left( 0 \right) + B\left( { - 1} \right)\left( { - 1 - 1} \right) + C\left( 0 \right) \cr & B = 1/2 \cr & {\text{if we set }}x = 3 \cr & 1 = A\left( 0 \right) + B\left( 0 \right) + C\left( 1 \right)\left( {1 + 1} \right) \cr & C = 1/2 \cr & \cr & {\text{then}} \cr & \frac{1}{{x\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{ - 1}}{x} + \frac{{1/2}}{{x + 1}} + \frac{{1/2}}{{x - 1}} \cr & \cr & \int {\frac{{dx}}{{x\left( {{x^2} - 1} \right)}}} = \int {\left( {\frac{{ - 1}}{x} + \frac{{1/2}}{{x + 1}} + \frac{{1/2}}{{x - 1}}} \right)dx} \cr & {\text{integrating}} \cr & = - \ln \left| x \right| + \frac{1}{2}\ln \left| {x + 1} \right| - \frac{1}{2}\ln \left| {x - 1} \right| + C \cr & = - \ln \left| x \right| + \frac{1}{2}\ln \left| {\frac{{x + 1}}{{x - 1}}} \right| + C \cr} $$