# Find inverse laplace for: f(s)=(1)/((s−2)(s+2)^2)

Find inverse laplace for:
$f\left(s\right)=\frac{1}{\left(s-2\right)\left(s+2{\right)}^{2}}$
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Baluttor7
The inverse Laplace Transform is given by
$f\left(t\right)=\frac{1}{2\pi i}{\int }_{{2}^{+}-i\mathrm{\infty }}^{{2}^{+}+i\mathrm{\infty }}\frac{1}{\left(s-2\right)\left(s+2{\right)}^{2}}{e}^{st}\phantom{\rule{thinmathspace}{0ex}}ds$
For t>0, we close the contour in the left-half plane $s<{2}^{+}$. Note that $F\left(s\right){e}^{st}$ has a simple pole at s=2 and a second-order pole at s=−2.
The associated residues of $F\left(s\right){e}^{st}$ are $\frac{{e}^{2t}}{16}$ at $s=2$ and $-\frac{1}{4}t{e}^{-2t}-\frac{1}{16}{e}^{-2t}$ at $s=-2$
Hence, we find that for t>0
$f\left(t\right)=\frac{1}{16}{e}^{2t}-\frac{1}{16}{e}^{-2t}-\frac{1}{4}t{e}^{-2t}$
For t<0, we close the contour in the right-half plane. Since $F\left(s\right){e}^{st}$ is analytic in that plane, f(t)=0 for t<0.