Chapter 6.9, Problem 67ES

The integration formulas for $1/\sqrt{u^2-a^2}$ in Theorem $\mathrm{6.9.6}$ are valid for $u>a$ . Show that the following formula is valid for $u<-a$ :

${\displaystyle \int \frac{du}{\sqrt{u^2-a^2}}}=-{\cosh }^{-1}\left(-\frac{u}{a}\right)+C$ or $\ln \left|u+\sqrt{u^2-a^2}\right|+C$