## Calculus (3rd Edition)

$$\frac{1}{2} \sec^{-1}x^2+c.$$
Let $u=x^2$, then $du=2xdx$ and hence we have $$\int \frac{xdx}{x\sqrt{x^4-1}}=\frac{1}{2}\int \frac{du}{u\sqrt{u^2-1}}=\frac{1}{2} \sec^{-1}u+c\\ =\frac{1}{2} \sec^{-1}x^2+c.$$