Find the inverse laplace transform of 1/((s^2+1)(s^2+2s+3)).

Question

zepdrix · Answer

Hmm I'm not really sure how to approach this one.
Could we do this maybe?$$\large s^2+2s+3 \qquad=\qquad (s^2+2s+1)+2 \qquad=\qquad (s+1)^2+2$$

Then I'm not sure if I'm setting up these partial fractions up correctly, or if they'll even work. Just an idea though:$$\large \frac{1}{(s^2+1)\left[(s+1)^2+2\right]}=\frac{As+B}{s^2+1}+\frac{C(s+1)+D}{(s+1)^2+2}$$

anonymous · Answer

Let me try.

anonymous · Answer

How do you solve for A, B, C, D?

zepdrix · Answer

Expand eeeeeeeverything out. :[
Equate like terms.

loser66 · Answer

back to your previous post to get it, girl

zepdrix · Answer

Oh did you mean to type:$$\Large s^2+3s+2$$Idealist?

Because that's quite a bit different than what you have above,$$\Large s^2+2s+3$$:(

anonymous · Answer

@zepdrix
$$\frac{1}{(s^2+1)\left[(s+1)^2+2\right]}=\frac{As+B}{s^2+1}+\frac{C\color{red}s+D}{(s+1)^2+2}$$
Using $C(s+1)$ gives you $Cs+C$, which mucks up the next few steps. You'll find that $D=0$, whereas using $Cs$ gives you $D=\dfrac{1}{4}$.