Partial fractions for inverse laplace transform

Question

I have the following function for which I need to find the inverse laplace transform:

$$\frac1{s(s^2+1)^2}$$

Am I correct in saying the partial fraction is:

$$\frac1{s(s^2+1)^2}=\frac{A}{s}+\frac{Bs+C}{s^2+1}+\frac{Ds+E}{(s^2+1)^2}$$

Answer 1

Your guess is correct. The coefficients are...SPOILER ALERT...

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$$A=1,$$ $$B=-1,$$ $$C=E=0,$$ $$D=-1.$$

Answer 2

Another way to evaluate the ILT when there are just poles (i.e., no branch points like roots and logarithms) is to apply the residue theorem. In this case, the ILT is simply the sum of the residues at the poles of the LT. That is,

$$\begin{align}f(t) &= \operatorname*{Res}_{s=0} \frac{e^{s t}}{s(1+s^2)^2}+ \operatorname*{Res}_{s=i} \frac{e^{s t}}{s(1+s^2)^2}+\operatorname*{Res}_{s=-i} \frac{e^{s t}}{s(1+s^2)^2}\\ &= \frac{e^{(0) t}}{(0^2+1)^2}+\left[\frac{d}{ds}\frac{e^{s t}}{s(s+i)^2}\right]_{s=i} +\left[\frac{d}{ds}\frac{e^{s t}}{s(s-i)^2}\right]_{s=-i}\\ &= 1+\left[ \frac{e^{s t} ((s+i) s t-3 s-i)}{s^2 (s+i)^3}\right]_{s=i}+\left [\frac{e^{s t} ((s-i) s t-3 s+i)}{s^2 (s-i)^3} \right ]_{s=-i}\\ &=1+\left[-\frac{1}{8} i e^{i t} (-2 t-4 i)\right]+\left[\frac{1}{8} i e^{-i t} (-2 t+4 i)\right]\\&=1-\cos{t}-\frac12 t \sin{t}\end{align}$$