Inverse Laplace transformation of $\frac{s^2}{(s^2+1)^2}$ I am a bit lost how to compute the inverse Laplace transformation of $$ \frac {s^2}{(s^2+1)^2}$$ I think it will be some combination of sine and cosine (oscillation-like), however I ran into some weird expressions when I tried to expand the expression as $$ \frac {s}{(s^2+1)}  \frac {s}{(s^2+1)}$$
Could someone please give some suggestions? Thanks in advance!
 A: $$
F(s)=\frac{s^2}{(s^2 + 1)^2} = \frac{s^2 + 1}{(s^2 + 1)^2} - \frac{1}{(s^2 + 1)^2}$$
$$
F(s)= \frac 1{s^2+1} +\dfrac {1}{2s}\dfrac {d}{ds}\frac 1{(s^2 + 1)}.
$$
Apply inverse Laplace Transform:
$$f(t)=\sin t -\dfrac 12 \int_0^t 1  \times \tau \sin \tau \ d\tau$$
$$f(t)=\dfrac 12 (\sin t +t \cos t)$$
A: The original problem was for $\frac{s^2}{(s^2 + 1)^2}$
Consider the Laplace transforms of $\sin(a t)$ and $t \, \cos(a t)$ which are
\begin{align}
\sin(a t) &\doteqdot \frac{a}{s^2 + a^2} = \frac{a \, s^2 + a^3}{(s^2 + a^2)^{2}} \\
t \, \cos(a t) &\doteqdot \frac{s^2 - a^2}{(s^2 + a^2)^{2}}.
\end{align}
Now one can notice that
$$ \frac{1}{2 a} \, \left( \sin(a t) + a \, t \, \cos(a t) \right) \doteqdot \frac{s^2}{(s^2 + a^2)^{2}}. $$
A second method is to use the convolution theorem,
$$ \bar{f}(s) \, \bar{g}(s) \doteqdot \int_{0}^{t} f(t-u) \, g(u) \, du $$
where $\bar{f}(s)$ is the transform of $f(t)$. Since
$$ \cos(a t) \doteqdot \frac{s}{s^2 + a^2}$$
then
\begin{align}
\frac{s^2}{(s^2 + a^2)^2} &\doteqdot \int_{0}^{t} \cos(a u) \, \cos(at - au) \, du \\
&\doteqdot \left[\frac{2 a t \, \cos(a t) - \sin(a (t - 2 u))}{4 a} \right]_{0}^{t} \\
&\doteqdot \frac{1}{2 a} \, \left( \sin(a t) + a \, t \, \cos(a t) \right).
\end{align}
Modified form $\frac{s^3}{(s^2+1)^2}$
By using :
\begin{align}
\cos(a t) - a t \, \sin(a t) &\doteqdot \frac{s \, (s^2 - a^2)}{(s^2 + a^2)^2} \\
\cos(a t) + a t \, \sin(a t) &\doteqdot \frac{s \, (s^2 + 3 a^2)}{(s^2 + a^2)^2}
\end{align}
then it can be shown that
$$ \frac{1}{2} \, ( 2 \, \cos(a t) - a \, t \, \sin(a t) ) \doteqdot \frac{s^3}{(s^2 + a^2)^2}. $$
