## Calculus: Early Transcendentals (2nd Edition)

$$3\ln \left| {x - 1} \right| - 3\ln \left| {x + 1} \right| + C$$
\eqalign{ & \int {\frac{6}{{{x^2} - 1}}} dx \cr & {\text{partial fraction decomposition}} \cr & \frac{6}{{{x^2} - 1}} = \frac{6}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \cr & \frac{6}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} \cr & 6 = A\left( {x + 1} \right) + B\left( {x - 1} \right) \cr & {\text{for }}x = 1 \cr & 6 = A\left( {1 + 1} \right) + B\left( {1 - 1} \right) \cr & 6 = 2A \cr & 3 = A \cr & {\text{for }}x = - 2 \cr & 6 = A\left( { - 1 + 1} \right) + B\left( { - 1 - 1} \right) \cr & 6 = - 2B \cr & - 3 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 1}} + \frac{B}{{x + 1}} = \frac{3}{{x - 1}} - \frac{3}{{x + 1}} \cr & \int {\frac{6}{{{x^2} - 1}}} dx = \int {\left( {\frac{3}{{x - 1}} - \frac{3}{{x + 1}}} \right)} dx \cr & {\text{sum rule}} \cr & = \int {\frac{3}{{x - 1}}} dx - \int {\frac{3}{{x + 1}}} dx \cr & {\text{integrating}} \cr & = 3\ln \left| {x - 1} \right| - 3\ln \left| {x + 1} \right| + C \cr}