# How to find inverse laplace of ((s^2−6s+7))/((s^2−4s+5)^2)?

How to find inverse laplace of $\frac{\left({s}^{2}-6s+7\right)}{{\left({s}^{2}-4s+5\right)}^{2}}$?
i tried partial frictions but i get $\frac{2-2s}{{\left({s}^{2}-4s+5\right)}^{2}}+\frac{1}{{s}^{2}-4s+5}$
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berzamauw
Just another way:
${\mathcal{L}}_{s}^{-1}\left[\frac{{s}^{2}-6s+7}{{\left({s}^{2}-4s+5\right)}^{2}}\right]\left(t\right)=\phantom{\rule{0ex}{0ex}}-t\left({\mathcal{L}}_{s}^{-1}\left[\int \frac{{s}^{2}-6s+7}{{\left({s}^{2}-4s+5\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}ds\right]\left(t\right)\right)=\phantom{\rule{0ex}{0ex}}-t\left({\mathcal{L}}_{s}^{-1}\left[\frac{3-s}{5-4s+{s}^{2}}\right]\left(t\right)\right)$
Using Heaviside's expansion formula for: ${\mathcal{L}}_{s}^{-1}\left[\frac{3-s}{5-4s+{s}^{2}}\right]\left(t\right)$ we have:
$P\left(s\right)=3-s$
$Q\left(s\right)=5-4s+{s}^{2}$
${Q}^{\prime }\left(s\right)=-4+2s$
solve roots for:Q(s)
$\alpha \text{1}=2-i$ and $\alpha \text{2}=2-i$
$\frac{P\left(\alpha \text{1}\right)\mathrm{exp}\left(t\alpha \text{1}\right)}{{Q}^{\prime }\left(\alpha \text{1}\right)}+\frac{P\left(\alpha \text{2}\right)\mathrm{exp}\left(t\alpha \text{2}\right)}{{Q}^{\prime }\left(\alpha \text{2}\right)}=\frac{P\left(2-i\right)\mathrm{exp}\left(t\left(2-i\right)\right)}{{Q}^{\prime }\left(2-i\right)}+\frac{P\left(2+i\right)\mathrm{exp}\left(t\left(2+i\right)\right)}{{Q}^{\prime }\left(2+i\right)}=\left(-\frac{1}{2}+\frac{i}{2}\right)\mathrm{exp}\left(t\left(2-i\right)\right)+\left(-\frac{1}{2}-\frac{i}{2}\right)\mathrm{exp}\left(t\left(2+i\right)\right)={e}^{2t}\left(-\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)\right)$
and then:
$-t\left({\mathcal{L}}_{s}^{-1}\left[\frac{3-s}{5-4s+{s}^{2}}\right]\left(t\right)\right)=-t{e}^{2t}\left(-\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)\right)={e}^{2t}t\left(\mathrm{cos}\left(t\right)-\mathrm{sin}\left(t\right)\right)$