Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 431: 63

$-\displaystyle \frac{\ln 3}{2}$

Substitute into the relevant formula: $\displaystyle \tanh^{-1}(-\frac{1}{2})=\frac{1}{2} \displaystyle \ln(\frac{1+(-\frac{1}{2})}{1-(-\frac{1}{2})})$ $=\displaystyle \frac{1}{2} \displaystyle \ln(\frac{\frac{1}{2}}{\frac{3}{2}})$ $=\displaystyle \frac{1}{2} \ln 3^{-1}$ $=-\displaystyle \frac{\ln 3}{2}$