Answer
$$\frac{1}{2}\ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{1}{2}{x^2} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}dx}}{{{x^2} - 1}}} \cr
& {\text{Use the trigonometric substitutions:}} \cr
& {\text{ }}x = \sec \theta,{\text{ }}dt = \sec \theta \tan \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{{x^3}dx}}{{{x^2} - 1}}} = \int {\frac{{{{\sec }^3}\theta \left( {\sec \theta \tan \theta d\theta } \right)}}{{{{\sec }^2}\theta - 1}}} \cr
& = \int {\frac{{{{\sec }^4}\theta \tan \theta }}{{{{\sec }^2}\theta - 1}}d\theta } \cr
& {\text{Use the fundamental identity }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& = \int {\frac{{{{\sec }^4}\theta \tan \theta }}{{{{\tan }^2}\theta }}d\theta } \cr
& = \int {\frac{{{{\sec }^4}\theta }}{{\tan \theta }}d\theta } \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta }}{{\sec }^2}\theta d\theta } \cr
& = \int {\left( {\frac{{1 + {{\tan }^2}\theta }}{{\tan \theta }}} \right){{\sec }^2}\theta d\theta } \cr
& = \int {\left( {\frac{1}{{\tan \theta }} + \tan \theta } \right){{\sec }^2}\theta d\theta } \cr
& {\text{Integrate }} \cr
& = \ln \left| {\tan \theta } \right| + \frac{{{{\tan }^2}\theta }}{2} + C \cr
& {\text{Write in terms of }}x,\tan \theta = (\sqrt {{x^2} - 1}) \cr
& = \ln \left| {\sqrt {{x^2} - 1} } \right| +
\frac{{\left({\sqrt {{x^2} - 1}}\right)}^2}{2} + C \cr
& = \ln \left| {\sqrt {{x^2} - 1} } \right| + \frac{{{x^2} - 1}}{2} + C \cr
& = \frac{1}{2}\ln \left| {{{x^2} - 1} } \right| + \frac{1}{2}{x^2} - \frac{1}{2} + C \cr
& {\text{Combine constants}} \cr
& = \frac{1}{2}\ln \left| {{{x^2} - 1} } \right| + \frac{1}{2}{x^2} + C \cr} $$
