I am really stumpted on this problem and can't seem to figure out where to go from where I am. Can anyone give me some advice or hint where I should do next? Here is the problem: $$L^{-1}(\frac{8s+8}{(s^2 + 2s + 5)^2})$$ So what I did was this: $$L^{-1}(\frac{8s+8}{(s^2 + 2s + 5)^2})=L^{-1}(-4(-\frac{d}{ds})(\frac{1}{s^2+2s+5}))$$ Then by completing the square and simplying the negatives I got this: $$=L^{-1}(4\frac{d}{ds}(\frac{1}{(s+1)^2 + 4}))$$ Simplying again I get: $$=4tL^{-1}(\frac{1}{(s+1)^2 + 4})$$ And this is where I get stuck.
One: I am not sure if what I have done up until not is correct.
Two: I am not sure what inverse Laplace transform this would be or how to get it.
I think that I have to use the first shifting theorem in order to solve this inverse Laplace transform. This is what I think it should look like or something similar: $$=4te^{-t}L^{-1}(\frac{1}{s^2+4})$$ But this gives: $$=2te^{-t}sin(2t)$$ Which I know is not the answer. So can anyone give me some help or point me in the right direction, it would be greatly apprecaited.
Thanks!
$2 t e^{-t} \sin(2t)$ is correct. Why do you think that's not the answer? If you have been given a different answer, it may have been expressed in a different form (e.g. expanding $\sin(2t)$ as $2 \sin(t)\cos(t)$).