Finding the Inverse laplace transform $\log\frac{s+1}{s-1}$

Question

Can anyone give me some hint how can I find the inverse Laplace Transform of :
$$\textrm{log}\bigg\{\frac{s+1}{s-1}\bigg\}$$

Answer 1

The fastest way, IMHO, is to consider that the inverse Laplace transform of the derivative is just: $$\mathcal{L}^{-1}\left(\frac{2}{1-s^2}\right) = -2\sinh x,$$ since $\frac{2}{1-s^2}=\frac{1}{1-s}+\frac{1}{1+s}$, hence: $$\mathcal{L}^{-1}\left(\log\frac{s+1}{s-1}\right)=\frac{2\sinh x}{x}.$$