Answer
$$\sinh^{-1}\Big(-\frac{5}{12}\Big)=\ln\frac{2}{3}$$
Work Step by Step
We have $$\sinh^{-1}x=\ln(x+\sqrt{x^2+1})$$ for $x\in(-\infty,\infty)$
Therefore, $$\sinh^{-1}\Big(-\frac{5}{12}\Big)=\ln\Big(-\frac{5}{12}+\sqrt{\frac{25}{144}+1}\Big)$$ $$=\ln\Big(-\frac{5}{12}+\sqrt{\frac{169}{144}}\Big)$$ $$=\ln\Big(-\frac{5}{12}+\frac{13}{12}\Big)$$ $$=\ln\frac{2}{3}$$