Answer
$$\frac{1}{2}\ln \left( {{x^2} + \sqrt {{x^4} - 1} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {{x^4} - 1} }}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}\frac{{du}}{{2x}} = dx \cr
& {\text{Substitute}} \cr
& \int {\frac{x}{{\sqrt {{x^4} - 1} }}} dx = \int {\frac{x}{{\sqrt {{u^2} - 1} }}} \left( {\frac{1}{{2x}}} \right)du \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt {{u^2} - 1} }}} du \cr
& {\text{Using the rules of theorem 5}}{\text{.20 }}\left( {{\text{Page 389}}} \right) \cr
& = \frac{1}{2}\ln \left( {u + \sqrt {{u^2} - 1} } \right) + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{2}\ln \left( {{x^2} + \sqrt {{x^4} - 1} } \right) + C \cr} $$
