I need to find the inverse laplace of this : $$\frac{s+2}{s^2+2s+5}$$
I know that completing the square should help me to solve this so I get $$\frac{s+2}{(s+1)^2+4}$$
Then separating this equation i get $$\frac{s}{(s+1)^2+2^2} + \frac{2}{(s+1)^2+2^2}$$
I can find the inverse Laplace for the second part easily but I can't find it for $$\frac{s}{(s+1)^2+2^2}$$
What am I not understanding here?
On
Just another approach. The zeroes of $s^2+2s+5$ occur at $s=-1\pm 2i$, and: $$\text{Res}\left(\frac{s+2}{s^2+2s+5},s=-1\pm 2i\right)=\frac{1}{2}\pm \frac{i}{4},\tag{1}$$ so: $$ f(s) = \frac{s+2}{s^2+2s+5} = \frac{\frac{1}{2}+\frac{i}{4}}{s-(-1+2i)}+\frac{\frac{1}{2}-\frac{i}{4}}{s-(-1-2i)}\tag{2} $$ and since $\mathcal{L}^{-1}\left(\frac{1}{s-\alpha}\right)=e^{\alpha x}$ we have: $$\begin{eqnarray*}\mathcal{L}^{-1}(f(s)) &=& \left(\frac{1}{2}+\frac{i}{4}\right) e^{(-1+2i)x}+\left(\frac{1}{2}-\frac{i}{4}\right) e^{(-1-2i)x}\\&=&\color{red}{e^{-x}\cos(2x)+\frac{1}{2}e^{-x}\sin(2x)}.\tag{3}\end{eqnarray*}$$
I'm assuming there is a typo in the first equation. I will assume the second equation is right. Some standard Laplace transforms are $$\mathcal L[e^{-at}\cos(bt)]=\frac{s+a}{(s+a)^2+b^2},$$ and $$\mathcal L[e^{-at}\sin(bt)]=\frac{b}{(s+a)^2+b^2}.$$ You can write your Laplace transform as $$\mathcal L[f]=\frac{s+2}{(s+1)^2+2^2}\\=\frac{s+1}{(s+1)^2+2^2}+\frac{1}{(s+1)^2+2^2}\\=\frac{s+1}{(s+1)^2+2^2}+\frac{1}{2}\frac{2}{(s+1)^2+2^2}.$$ Thus, $$f(t)=\frac{1}{2}e^{-t}\sin(2t)+e^{-t}\cos(2t).$$