I've been having trouble with understanding how to properly do partial-fraction decomposition on the inverse Laplace transform of $$ F(s) = \frac{s+1}{(s^2+1)^2} $$ I tried using the complex factors $$ \frac{s+1}{(s^2+1)^2}=\frac{As+B}{(s-i)^2}+\frac{Cs+D}{(s+i)^2} $$ Finding $$ A = -\frac{i}{4} \\ B = -\frac{1}{2}-\frac{i}{4} \\ C = \frac{i}{4} \\ D = -\frac{1}{2}+\frac{i}{4} $$ But I have no clue how to proceed from here as I don't see a way to match this to any common laplace transforms.
I also tried this decomposition $$ \frac{s+1}{(s^2+1)^2}=\frac{As+B}{s^2+1}+\frac{Cs+D}{(s^2+1)^2} $$ Here I get $$ A = 0 \\ B = 0 \\ C = 1 \\ D = 1 $$ Which simply brings me back to square 1. If I look for what my course book gives as the general form I should arrive at for this problem, I find this: $$ f(t) = ((At+B)sin(t)+(Ct+D)cos(t))\theta(t) $$ where $\theta(t)$ is the Heaviside step function.
How can I manipulate the original $F(s)$ to see this general form of $f(t)$?
I hope this gives you help otherwise, tell me to delete it. We know that: $$ \mathcal{L}(t^nf(t))=(-1)^n\dfrac{d^nF}{ds^n} $$ where $n$ is a positive integer. Use this fact and see that $\left(\dfrac{-0.5}{s^2+1}\right)'=\dfrac{s}{(1+s^2)^2}$. Moreover, apply the method pointed here for the rest to find the inverse Laplace of $\dfrac{1}{(1+s^2)^2}$. I think you find the way!