I thought maybe you could fist solve $\frac{(3s^2-s+2)}{(s-1)(s^2+1)}$ using partial fractions and later solve the $e^{-s}$ separately as it is a $d(t-1)$ (Dirac delta function).
As you solve the partial fraction you will obtain as an answer $2e^t+\cos(t)$, I just can't figure out if at the end I just multiply my answers? I don't know what to do.
Not even sure if this is the correct way to do solve this.
I prefer the method of residues to partial fractions. If you recall, the residue of a simple pole $s_0$ of a function $f(s)$ is
$$\lim_{s \rightarrow s_0} [(s-s_0) f(s)]$$
Here
$$f(s) = \frac{3 s^2-s+2}{(s-1)(s^2+1)} e^{s t}$$
Therefore are poles at $s=1$ and $s=\pm i$. The sum of the residues at these poles are
$$2 e^{t} + \frac{-1-i}{(-1+i)(2 i)} e^{i t} + \frac{-1+i}{(-1-i)(-2 i)} e^{-i t} = 2 e^t + \cos{t}$$
and that is the ILT.