 # How to compute the inverse Laplace transformation of (s^2)/((s^2+1)^2) metal1fc 2022-09-12 Answered
How to compute the inverse Laplace transformation of $\frac{{s}^{2}}{\left({s}^{2}+1{\right)}^{2}}$
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$F\left(s\right)=\frac{{s}^{2}}{\left({s}^{2}+1{\right)}^{2}}=\frac{{s}^{2}+1}{\left({s}^{2}+1{\right)}^{2}}-\frac{1}{\left({s}^{2}+1{\right)}^{2}}$
$F\left(s\right)=\frac{1}{{s}^{2}+1}+\frac{1}{2s}\frac{d}{ds}\frac{1}{\left({s}^{2}+1\right)}.$
Apply inverse Laplace Transform:

$f\left(t\right)=\frac{1}{2}\left(\mathrm{sin}t+t\mathrm{cos}t\right)$
###### Not exactly what you’re looking for? Skye Vazquez
Consider the Laplace transforms of $\mathrm{sin}\left(at\right)$ and $t\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(at\right)$ which are
$\begin{array}{rl}\mathrm{sin}\left(at\right)& \doteqdot \frac{a}{{s}^{2}+{a}^{2}}=\frac{a\phantom{\rule{thinmathspace}{0ex}}{s}^{2}+{a}^{3}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\\ t\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(at\right)& \doteqdot \frac{{s}^{2}-{a}^{2}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}.\end{array}$
Now one can notice that
$\overline{f}\left(s\right)\phantom{\rule{thinmathspace}{0ex}}\overline{g}\left(s\right)\doteqdot {\int }_{0}^{t}f\left(t-u\right)\phantom{\rule{thinmathspace}{0ex}}g\left(u\right)\phantom{\rule{thinmathspace}{0ex}}du$
A second method is to use the convolution theorem,
f¯(s)g¯(s)≑∫t0f(t−u)g(u)du
where $\overline{f}\left(s\right)$ is the transform of $f\left(t\right)$. Since
$\mathrm{cos}\left(at\right)\doteqdot \frac{s}{{s}^{2}+{a}^{2}}$
then
$\begin{array}{rl}\frac{{s}^{2}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}& \doteqdot {\int }_{0}^{t}\mathrm{cos}\left(au\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(at-au\right)\phantom{\rule{thinmathspace}{0ex}}du\\ & \doteqdot {\left[\frac{2at\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(at\right)-\mathrm{sin}\left(a\left(t-2u\right)\right)}{4a}\right]}_{0}^{t}\\ & \doteqdot \frac{1}{2a}\phantom{\rule{thinmathspace}{0ex}}\left(\mathrm{sin}\left(at\right)+a\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(at\right)\right).\end{array}$
Modified form $\frac{{s}^{3}}{\left({s}^{2}+1{\right)}^{2}}$
By using :
$\begin{array}{rl}\mathrm{cos}\left(at\right)-at\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(at\right)& \doteqdot \frac{s\phantom{\rule{thinmathspace}{0ex}}\left({s}^{2}-{a}^{2}\right)}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\\ \mathrm{cos}\left(at\right)+at\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(at\right)& \doteqdot \frac{s\phantom{\rule{thinmathspace}{0ex}}\left({s}^{2}+3{a}^{2}\right)}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\end{array}$
then it can be shown that
$\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\left(2\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(at\right)-a\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(at\right)\right)\doteqdot \frac{{s}^{3}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}.$