How do I calculate this inverse Laplace Transform
I understand you have
\begin{equation}
R(s) = \frac{40}{s^2+10^2}
\end{equation}
So
\begin{equation}
Y(s) = FT_6(s)R(s)=\frac{4}{(s^2+10^2)(s^2+s+0.1)}
\end{equation}
Then you proceed by taking partial fractions of Y(s)
\begin{equation}
Y(s) = \frac{A+sB}{(s^2+10^2)} +\frac{C+sD}{(s^2+s+0.1)}
\end{equation}
Then you expand the term
\begin{equation}
\frac{C+sD}{s^2+s+0.1} = \frac{C+sD}{(s+\frac{1}{2})^2-0.15} = 
D\frac{s+\frac{1}{2}}{(s+\frac{1}{2})^2-0.15} +
(C-\frac{D}{2})\frac{1}{(s+\frac{1}{2})^2-0.15}
\end{equation}
Now you can use a table (keeping in mind the table items 7,8,21 and 22).