Having a little trouble solving this fractional inverse Laplace were the den. is a irreducible repeated factor
$$F(s) = \frac{3s+8}{(s^2+2s+20)^2}$$
tried to apply partial fractions to it and i just get to the same result as F(S), rewriting the den as $s^4+4 s^3+44 s^2+80 s+400$ didn't help much either
Its driving me crazy its the last one i need to complete a test. In my class we haven't got to convolution yet, but a friend told me that i should try that, id appreciate if you guys help me with that by any possible method.
EDIT:
got to $$F(s) = \frac{3s+8}{(s^2+2s+20)^2}= \frac{3(s-1)}{((s+1)^2+19)^2}+ \frac {11}{((s+1)^2+19)^2}$$
using lineality:
$$ 3\times\left(\\{L^{-1}} \frac{(s-1)}{((s+1)^2+19)^2}\right)+11\times\left(\\{L^{-1}} \frac{1}{((s+1)^2+19)^2}\right) $$
im kinda new at this, i dont know what happens that the ecuation doesnt stay put in one line, but i think its understanadable
Using this, $$L^{-1}F'(s)=-t\cdot f(t)$$
Let $$\dfrac{d\left(\dfrac{as+b}{s^2+2s+20}\right)}{ds}=\dfrac{3s+8}{(s^2+2s+20)^2}$$
Find $a,b$
Now use this, $$A\cdot L^{-1}\dfrac{s-a}{(s-a)^2+b^2}+B\cdot L^{-1}\dfrac b{(s-a)^2+b^2}=A\cdot e^{at}\cos bt+B\cdot e^{at}\sin bt$$