Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises - Page 550: 87

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

$$\eqalign{ & \int {\frac{{dx}}{{{x^2} - 1}}} ,{\text{ }}x > 1 \cr & \cr & *{\text{Using partial fraction decomposition}} \cr & \frac{1}{{{x^2} - 1}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} \cr & 1 = A\left( {x + 1} \right) + B\left( {x - 1} \right) \cr & x = 1 \to A = \frac{1}{2} \cr & x = - 1 \to B = - \frac{1}{2} \cr & \frac{1}{{{x^2} - 1}} = \frac{{1/2}}{{x - 1}} - \frac{{1/2}}{{x + 1}} \cr & {\text{Therefore,}} \cr & \int {\frac{{dx}}{{{x^2} - 1}}} = \int {\left( {\frac{{1/2}}{{x - 1}} - \frac{{1/2}}{{x + 1}}} \right)} dx \cr & {\text{Integrate}} \cr & \int {\frac{{dx}}{{{x^2} - 1}}} = \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + C \cr & \int {\frac{{dx}}{{{x^2} - 1}}} = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C \cr & \cr & *{\text{Using trigonometric substitution}} \cr & \int {\frac{{dx}}{{{x^2} - 1}}} \cr & {\text{Let }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr & \int {\frac{{dx}}{{{x^2} - 1}}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {\sec \theta } \right)}^2} - 1}}} \cr & = \int {\frac{{\sec \theta \tan \theta }}{{{{\sec }^2}\theta - 1}}} d\theta \cr & = \int {\frac{{\sec \theta \tan \theta }}{{{{\tan }^2}\theta }}} d\theta \cr & = \int {\frac{{\sec \theta }}{{\tan \theta }}} d\theta = \int {\frac{1}{{\sin \theta }}} d\theta \cr & = \int {\csc \theta } d\theta \cr & {\text{Integrate}} \cr & = - \ln \left| {\csc \theta + \cot \theta } \right| + C \cr & {\text{Write in terms of }}x \cr & = - \ln \left| {\frac{x}{{\sqrt {{x^2} - 1} }} + \frac{1}{{\sqrt {{x^2} - 1} }}} \right| + C \cr & = - \ln \left| {\frac{{x + 1}}{{\sqrt {{x^2} - 1} }}} \right| + C \cr & = \ln \left| {\frac{{\sqrt {{x^2} - 1} }}{{x + 1}}} \right| + C \cr & = \ln \sqrt {{x^2} - 1} - \ln \left| {x + 1} \right| + C \cr & = \frac{1}{2}\ln \left| {\left( {x + 1} \right)\left( {x - 1} \right)} \right| - \ln \left| {x + 1} \right| + C \cr & = \frac{1}{2}\ln \left| {x + 1} \right| + \frac{1}{2}\ln \left| {x - 1} \right| - \ln \left| {x + 1} \right| + C \cr & = \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + C \cr & = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C \cr} $$