Partial fraction expansion of $\frac{1}{(s+1)^{2}(s-1)(s+5)}$ I'm seeking a partial-fraction expansion of $A=\frac{1}{(s+1)^{2}(s-1)(s+5)}.$
I was solve equation differential using Laplace transform, but I need use partial fraction of $A$. 
 A: The partial fraction expansion is:
$$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{1}{24(s-1)} -\frac{1}{32(s+1)} - \frac{1}{8(s+1)^2} - \frac{1}{96(s+5)} $$
The fraction expansion can be expressed as:
$$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{D}{s+5}$$
$$ \implies A(s+1)^2(s+5) + B(s+1)(s-1)(s+5) + C(s-1)(s+5) + D(s-1)(s+1)^2 = 1$$
Using $s=1$, we get:
$$24A  = 1 \implies A = \frac{1}{24}$$
Using $s=-1$, we get:
$$-8C = 1 \implies C = -\frac{1}{8}$$
Using $s = -5$, we get:
$$-96D = 1 \implies D = -\frac{1}{96}$$
For $B$, none of these will work. So we can choose something that makes life easier. Using $s=0$, we get
$$5A -5B -5C -D = 1 $$
$$\implies \frac{5}{24} - 5B  + \frac{5}{8} +\frac{1}{96} = 1$$
$$\implies B = -\frac{1}{5}\left(1 - \frac{20}{96} - \frac{60}{96}  - \frac{1}{96}\right) = -\frac{1}{32}$$
A: Guide:
Let $$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{B}{s+1}+\frac{C}{(s+1)^2}+\frac{D}{s-1}+\frac{E}{(s+5)}$$
$$1=B(s-1)(s+1)(s+5)+C(s-1)(s+5)+D(s+1)(s+5)+E(s+1)^2(s-1)$$
By letting $s=1, -1, 5$, you should be able to solve for $C,D,E$. Now you just have to solve for $B$.
