sinh−1(2)=ln(2+22+1)=ln(2+5)≈1.4436sinh^{-1}(2) = ln(2 + \sqrt{2^2+1})= ln(2 + \sqrt{5})\approx 1.4436sinh−1(2)=ln(2+22+1)=ln(2+5)≈1.4436 tanh−1(1/4)=12ln(1+1/41−1/4)=12ln(5/43/4)=12ln(53)≈0.2554tanh^{-1}(1/4) =\frac{1}{2} ln\left(\frac{1 + 1 / 4}{1-1 / 4}\right)=\frac{1}{2}ln\left(\frac{5/4}{3/4}\right)=\frac{1}{2}ln\left(\frac{5}{3}\right)\approx 0.2554tanh−1(1/4)=21ln(1−1/41+1/4)=21ln(3/45/4)=21ln(35)≈0.2554