Inverse Laplace with $\ln$ How can I compute the inverse Laplace of  
1) $\ln\left(\dfrac{s+1}{s-1}\right)$
2) $\ln\left(\dfrac{s-1}{s}\right)$.
Can someone please hep me to do these two problems
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\large\tt\mbox{With}\quad \gamma > 1:}$
\begin{align}&\color{#66f}{\large%
\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
\ln\pars{s + 1 \over s - 1}\,\expo{st}\,{\dd s \over 2\pi\ic}}
=\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
\int_{-1}^{1}{\dd x \over x + s}\,\expo{st}\,{\dd s \over 2\pi\ic}
\\[3mm]&=\int_{-1}^{1}\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\expo{st} \over s + x}\,{\dd s \over 2\pi\ic}\,\dd x
=\int_{-1}^{1}\expo{-xt}\,\dd x
={\expo{-t} - \expo{t}  \over -t} = \color{#66f}{\large{2\sinh\pars{t} \over t}}
\end{align}

The other one can be evaluated in a similar way.

A: Given
\begin{align}
\mathcal{L}\{f(t); s\} = \int_{0}^{\infty} e^{-st} f(t) \ dt
\end{align}
then the inverse Laplace transforms are given by the following
\begin{align}
\mathcal{L}\{ \frac{2 \sinh(at)}{t}; s\} = \ln\left(\frac{s+a}{s-a}\right)
\end{align}
and
\begin{align}
\mathcal{L}\{ \frac{1 - e^{at}}{t}; s\} = \ln\left(\frac{s-a}{s}\right).
\end{align}
The process to find the values given above is to use the relation
\begin{align}
\mathcal{L}\{ \frac{f(t)}{t}; s\} = \int_{s}^{\infty} g(u) \ du
\end{align}
where 
\begin{align}
g(s) = \int_{0}^{\infty} e^{- st} f(t) \ dt.
\end{align}
A: Podes usar algunas propiedades de las transformadas. 
[You can use some properties of the transforms]
Ej; [Example]
$$((-1)^n) (t^n) f(t)=F^n(s)$$
Donde f(t) es la antitransformada de $F^n(s)$. 
[Where $f(t)$ is in the inverse transform of $F^{n}(s)$.
