Use Laplace Transforms to give a solution formula for the initial value problem
$$u''(t)+2u'(t)+2u(t)=f(t), \ t\geq 0, \ u(0)=1, \ u'(0)=0. \tag1$$
where $f$ is a function such that $|f(t)|\leq Ate^B$ for some $A,B\in\mathbb{R}.$
Taking the LT of both sides I end upp with
$$\tilde{u}(s^2+2s+2)=\tilde{f}+s+2\implies\tilde{u}=\frac{\tilde{f}+s+2}{s^2+2s+2} = \frac{\tilde{f}+s+2}{(s+1)^2+1}, \tag2$$
where tilde denotes the laplace transform. I really have no idea how to proceed after this. Nowhere in my table can I see something that resembles the RHS in $(2)$ so I can't use any inversion formula.
EDIT 1:
Ok I actually got it now I think. Just need to split the last fraction up and complete the square of the denominator...
EDIT 2:
And I'm now stuck here:
$$\tilde{u}=\frac{\tilde{f}}{(s+1)+1}+\frac{s+1}{(s+1)+1}+\frac{1}{(s+1)+1}.$$
There is a formula that says that $L(e^{ct}f(t))=F(s-c),$ $c\in\mathbb{C.}$ But I don't see how it's applicable here...
$$ \bar u=\frac{\tilde{f}+s+2}{(s+1)^2+1} = \frac{\tilde{f}}{(s+1)^2+1}+\frac{s+2}{(s+1)^2+1} $$
then using the suitable inversion tables
$$ u(t) =((\cos t+\sin t)e^{-t}+f * ( e^{-t}\sin t))\phi(t) $$
where $\phi(t)$ is the unit step function and $\cdot * \cdot$ is the convolution operator