Find the inverse Laplace transform: $$x(t) = {L^{ - 1}}\left[ {\frac{1}{{{{({s^2} + 1)}^3}}}} \right]$$ with $x(t=0)=0$.
I did: $${\left[ {{\mathop{\rm R}\nolimits} {\rm{es}}\frac{{{e^{st}}{{(s - i)}^3}}}{{{{(s + i)}^3}{{(s - i)}^3}}},i} \right]_3} = \frac{1}{{2!}}\frac{{{d^2}}}{{d{s^2}}}\left[ {\frac{{{e^{st}}}}{{{{(s + i)}^3}}}} \right]$$ $$\begin{array}{l} g(s) = \frac{{{e^{st}}}}{{{{(s + i)}^3}}} \\ g' = \frac{{t{e^{st}}{{(s + i)}^3} - {e^{st}}3{{(s + i)}^2}}}{{{{(s + i)}^6}}} = \frac{{{e^{st}}(ts + ti - 3)}}{{{{(s + i)}^4}}}; \\ \end{array}$$ $$ g''= \frac{{{e^{st}}\left[ {{t^2}{s^2} + 2{t^2}is - 6ti - 6ts - {t^2} + 12} \right]}}{{{{(s + i)}^5}}}$$ $$g''(i) = \frac{1}{{32}}\left[ {4{t^2}\cos t - 12\cos t - 12t\sin t} \right] + i\frac{1}{{32}}\left[ {12t\cos t - 4{t^2}\sin t - 12\sin t} \right]$$ $$\begin{array}{l} \implies {\left[ {{\mathop{\rm R}\nolimits} {\rm{es}}\frac{{{e^{st}}{{(s - i)}^3}}}{{{{(s + i)}^3}{{(s - i)}^3}}},i} \right]_3} = \frac{1}{{2!}}\frac{{{d^2}}}{{d{s^2}}}\left[ {\frac{{{e^{st}}}}{{{{(s + i)}^3}}}} \right] \\ = \frac{1}{2}\left\{ {\frac{1}{{32}}\left[ {4{t^2}\cos t - 12\cos t - 12t\sin t} \right] + i\frac{1}{{32}}\left[ {12t\cos t - 4{t^2}\sin t - 12\sin t} \right]} \right\} \\ \implies \text{REAL}\left[ {{\mathop{\rm R}\nolimits} {\rm{es}}\frac{{{e^{st}}{{(s - i)}^3}}}{{{{(s + i)}^3}{{(s - i)}^3}}},i} \right] = \frac{1}{2}\frac{1}{{32}}\left[ {4{t^2}\cos t - 12\cos t - 12t\sin t} \right] \\ = \frac{1}{{16}}\left[ {{t^2}\cos t - 4\cos t - 4t\sin t} \right] \\ \implies x(t) = 2\text{REAL} = \frac{1}{8}\left[ {{t^2}\cos t - 4\cos t - 4t\sin t} \right] \\ \end{array}$$ However, when let $t=0$, $x(0)\neq 0$.