Inverse Laplace transform of $\frac{2s+1}{s^2 - 4s + 5}$ I approached this problem as follow: 


*

*Complete the square in the denominator and obtain $$(s-2)^2 + 1$$

*Now break the function into 2 parts. 
$$ \frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1}$$ 
Now the inverse laplace transform is straight forward.
$$e^{2t}(2cos(t) + sin(t))$$
However! I am wrong. The solution to this problem is actually 
$$e^{2t}(2cos(t) + 5*sin(t))$$
What is wrong with my approach?
Thanks.
 A: I think if you do the fraction as follows:
$$\frac{2s+1}{(s-2)^2 + 1}= \frac{2(s-2)+5}{(s-2)^2 + 1}=\frac{2(s-2)}{(s-2)^2 + 1}+\frac{5}{(s-2)^2 + 1}$$ and considering very useful fact:

$\mathcal{L}^{-1}(F(s-a))=e^{at}f(t)$

then we get:
$$\mathcal{L}^{-1}\left(\frac{2s+1}{(s-2)^2 + 1}\right)=2\mathcal{L}^{-1}\left(\frac{(s-2)}{(s-2)^2 + 1}\right)+5\mathcal{L}^{-1}\left(\frac{1}{(s-2)^2 + 1}\right)=2e^{2t}\cos t+5e^{2t}\sin t$$
A: Remember, the Laplace transform tables have the following kinds of terms
$$
\frac{s-a}{(s-a)^2+b^2},\frac{b}{(s-a)^2+b^2}
$$
Your term with an $s$ in the numerator has to look like $s-2$ in order to use that form.  So your problem is at this step:
$$
\frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = \frac{2(s-2)+4}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} =  2\frac{(s-2)}{(s-2)^2 + 1}+ 5\frac{1}{(s-2)^2 + 1}
$$ 
Basically, you have to have the same form for $s$ in the top and bottom, so you add and subtract $4$ from the numerator of that first term so that you can put it in terms of $(s-2)$.  Then the extra $4$ you added goes over to the other term to make a total of $5$, the coefficient of the $\sin$
A: $s^2-4 s+5$ has zeroes at $s_{\pm} = 2 \pm i$.  The ILT $f(t)$ is simply the sum of the residues of 
$$\frac{2 s+1}{s^2-4 s+5} e^{s t}$$
at these poles.  Then
$$f(t) = \frac{2 s_+ + 1}{2 s_+-4} e^{s_+ t} + \frac{2 s_- + 1}{2 s_--4} e^{s_- t}$$
Expanding this a bit:
$$f(t) = e^{2 t} \left [ \left (1-i \frac52 \right ) (\cos{t}+i \sin{t}) + \left(1+i \frac52 \right ) (\cos{t}-i \sin{t})  \right ]$$
Simplifying, I get
$$f(t) = e^{2 t} (2 \cos{t} + 5 \sin{t})$$
