Answer
$\begin{align*}
(a)& \quad 0.481 \\
(b)& \quad 0.347 \end{align*}$
Work Step by Step
Use Th. 5.19
$\displaystyle \coth^{-1}x=\frac{1}{2}\ln\frac{x+1}{x-1}\qquad $Domain: $(-\infty, -1)\cup(1, \infty)$
$\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1} x=\displaystyle \ln(\frac{1}{x}+\frac{\sqrt{1+x^{2}}}{|x|})\qquad $Domain: $(-\infty, 0)\cup(0, \infty)$
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(a)
$\displaystyle \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{-1}2=\ln(\frac{1+\sqrt{1+4}}{2})\approx 0.481$
(b)
$\displaystyle \coth^{-1}3=\frac{1}{2}\ln(\frac{3+1}{3-1})\approx 0.347$
