Difficult Inverse Laplace Transform I've had this question in my exam, which most of my batch mates couldn't solve it.The question by the way is the Laplace Transform inverse of  
$$\frac{\ln s}{(s+1)^2}$$
A Hint was also given, which includes the Laplace Transform of ln t.

Let $f(t)=\ln t$, then $F(s)=L\{f\}=-\frac{\gamma+\ln s}{s}$. So $\ln s=-sL\{f\}-\gamma$. Let $G(s)=\frac{s}{(s+1)^2}$ and then $g(t)=L^{-1}\{G\}=(t-1)e^{-t}$. Thus
$$ \frac{\ln s}{(s+1)^2}=-L\{f\}\frac{s}{(s+1)^2}-\frac{\gamma}{(s+1)^2}=-F(s)G(s)-\frac{\gamma}{(s+1)^2}. $$
Using
$$ F(s)G(s)=L\{\int_0^tf(\tau)g(t-\tau)d\tau\} $$
one has
\begin{eqnarray}
L^{-1}\{\frac{\ln s}{(s+1)^2}\}&=&-\int_0^tf(t)g(t-\tau)d\tau-\gamma L^{-1}\{\frac{1}{(s+1)^2}\} \\
&=&-\int_0^t\ln\tau(t-\tau-1)e^{-(t-\tau)}d\tau-\gamma te^{-t}\\
&=&e^{-t} (e^t-1-t\text{Chi}(t)-t\text{Shi}(t)).
\end{eqnarray}
Here $\text{Chi}(t), \text{Shi}(t)$ are CoshIntegaral and SinhIntegral.
