Find the inverse Laplace transform ccL^(-1){(s)/((s+1)^2−4)}.

Find the inverse Laplace transform ${\mathcal{L}}^{-1}\left\{\frac{s}{\left(s+1{\right)}^{2}-4}\right\}$
My textbook says that the solution is ${e}^{-t}\mathrm{cosh}\left(2t\right)-\frac{1}{2}{e}^{-t}\mathrm{sinh}\left(t\right)$
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Maggie Tanner
Hint:
$\frac{s}{\left(s+1{\right)}^{2}-{2}^{2}}=\frac{s}{\left(s-1\right)\left(s+3\right)}=\frac{1}{4\left(s-1\right)}+\frac{3}{4\left(s+3\right)},$
and
$\mathcal{L}\left({e}^{at}\right)=\frac{1}{s-a}.$
For the textbook solution, it is useful to write the given fraction as
$\frac{s}{\left(s+1{\right)}^{2}-{2}^{2}}=\frac{s+1-1}{\left(s+1{\right)}^{2}-{2}^{2}}=\frac{s+1}{\left(s+1{\right)}^{2}-{2}^{2}}-\frac{1}{2}\cdot \frac{2}{\left(s+1{\right)}^{2}-{2}^{2}},$
and note that
$\mathcal{L}\left({e}^{-at}\mathrm{cosh}\left(bt\right)\right)=\frac{s+a}{\left(s+a{\right)}^{2}-{b}^{2}},$
and
$\mathcal{L}\left({e}^{-at}\mathrm{sinh}\left(bt\right)\right)=\frac{b}{\left(s+a{\right)}^{2}-{b}^{2}}.$

Leroy Gray
${\mathcal{L}}^{-1}\left\{\frac{s}{\left(s+1{\right)}^{2}-4}\right\}$
$={\mathcal{L}}^{-1}\left\{\frac{s+1}{\left(s+1{\right)}^{2}-4}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{\left(s+1{\right)}^{2}-4}\right\}$
$={e}^{-t}\mathrm{cosh}\left(2t\right)-\frac{1}{2}{e}^{-t}\mathrm{sinh}\left(2t\right)$