# Need find a partial-fraction expansion of A=(1)/((s+1)^2 (s-1)(s+5))

Need find a partial-fraction expansion of $A=\frac{1}{\left(s+1{\right)}^{2}\left(s-1\right)\left(s+5\right)}.$
I was solve equation differential using Laplace transform, but I need use partial fraction of A.
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berzamauw
The partial fraction expansion is:
$\frac{1}{\left(s+1{\right)}^{2}\left(s-1\right)\left(s+5\right)}=\frac{1}{24\left(s-1\right)}-\frac{1}{32\left(s+1\right)}-\frac{1}{8\left(s+1{\right)}^{2}}-\frac{1}{96\left(s+5\right)}$
The fraction expansion can be expressed as:
$\frac{1}{\left(s+1{\right)}^{2}\left(s-1\right)\left(s+5\right)}=\frac{A}{s-1}+\frac{B}{s+1}+\frac{C}{\left(s+1{\right)}^{2}}+\frac{D}{s+5}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}A\left(s+1{\right)}^{2}\left(s+5\right)+B\left(s+1\right)\left(s-1\right)\left(s+5\right)+C\left(s-1\right)\left(s+5\right)+D\left(s-1\right)\left(s+1{\right)}^{2}=1$
Using $s=1$, we get:
$24A=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}A=\frac{1}{24}$
Using $s=-1$, we get:
$-8C=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}C=-\frac{1}{8}$
Using $s=-5$, we get:
$-96D=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}D=-\frac{1}{96}$
For B, none of these will work. So we can choose something that makes life easier. Using s=0, we get
$5A-5B-5C-D=1$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{5}{24}-5B+\frac{5}{8}+\frac{1}{96}=1$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}B=-\frac{1}{5}\left(1-\frac{20}{96}-\frac{60}{96}-\frac{1}{96}\right)=-\frac{1}{32}$