How to find the inverse Laplace Transforms? How do you find the inverse Laplace Transform of the following,
$$\frac{2 (s^2+4 s+5)^2+40}{(s^2+4 s+5)^2}$$
Separating them into complex coefficients is to long. 
 A: Let $F(s)$ denote the fraction in the post, hence $F(s)=2+40\frac1{(s^2+4s+5)^2}$. The $2$ part of $F(s)$ is the Laplace transform of twice the Dirac measure at $0$. The fraction $\frac1{s^2+4s+5}$ is a linear combination of $\frac1{s+2\pm\mathrm i}$ hence it is the Laplace transform of a linear combination of the functions $t\mapsto\exp(-(2\pm\mathrm i)t)$ on $t\geqslant0$, namely, the Laplace transform of $f$ where $f(t)=\mathrm e^{-2t}\sin(t)$ for every $t\geqslant0$. Thus, the fraction $\frac1{(s^2+4s+5)^2}$ is the Laplace transform of $f\ast f$, which might (or might not, check for yourself) be such that $(f\ast f)(t)=\frac12\mathrm e^{-2t}(\sin(t)-t\cos(t))$ for every $t\geqslant0$. 
Finally, the function $F$ is the Laplace transform of the measure $\nu=2\delta_0+40\mu$ where $\mu$ has density $f\ast f$ with respect to the Lebesgue measure on $t\geqslant0$. Note that $F$ is the Laplace transform of no function, only a measure can do, and one has
$$
F(s)=\int_0^{+\infty}\mathrm e^{-st}\mathrm d\nu(t).
$$
A: The problem reduces to $$2+\frac{40}{(s^2+4s+5)^2}$$
Let us denote the Laplace Transform of $f(t)$ as $L\left(f(t)\right)=F(s)$
and the inverse as $L^{-1}\left(F(s)\right)=f(t)$
For $L^{-1}\frac1{(s^2+4s+5)^2},$
$$\text{If }L\left(f(t)\right)=\frac{s-a}{\{(s-a)^2+b^2\}^2},$$
$$L\left(\frac {f(t)}t\right)=\int_s^\infty \frac{s-a}{\{(s-a)^2+b^2\}^2} ds$$
$$=\frac12\int_s^\infty \frac{2(s-a)}{\{(s-a)^2+b^2\}^2} ds=-\frac12\left(\frac1{(s-a)^2+b^2}\right)_s^\infty=\frac12\frac1{(s-a)^2+b^2}$$
$$\implies \frac {2f(t)}t=L^{-1}\left(\frac1{(s-a)^2+b^2}\right)=\frac{e^{at}\sin bt}{b} $$
$$\implies L^{-1}\frac{s-a}{\{(s-a)^2+b^2\}^2} =f(t)=\frac{te^{at}\sin bt}{2b}$$
$$a=0\implies  L^{-1}\frac s {\{s^2+b^2\}^2} =\frac{t\sin bt}{2b}$$
As  $L\{g(t)\}=G(s)\implies \int_0^tg(t)dt=L^{-1}\left(\frac{G(s)}s\right)$
$$L^{-1}\frac1{(s^2+b^2)^2}=L^{-1}\left(\frac1s\cdot \frac s{(s^2+b^2)^2}\right)=\int_0^t\frac{t\sin bt}{2b}dt=\frac{\sin bt-bt\cos bt}{2b^3}$$ (Integrating by parts)
Now using shifting property  $L\{g(t)\}=G(s)\implies L(e^{at}g(t))=G(s-a)$
So, $$L^{-1}\frac1{\{(s-a)^2+b^2\}^2}=\frac{e^{at}(\sin bt-bt\cos bt)}{2b^3}$$
Here $s^2+4s+5=(s+2)^2+1^2\implies a=-2,b=1$
A: I have done it, you can see if it's useful to you, I.m not sure so if you look out the mistake please fix it..
$\begin{array}{l}
 F(s) = \frac{{2{{({s^2} + 4s + 5)}^2} + 4}}{{{{({s^2} + 4s + 5)}^2}}} = 2 + \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}} \\ 
 {L^{ - 1}}{\rm{[2]  =  }}2\delta (t) \\ 
 {\rm{\{ }}\int_0^\infty  {2.{e^{ - st}}\delta (t)dt = 2} {\rm{.}}{{\rm{e}}^{ - s0}}{\rm{ = 2\} }} \\ 
 G(s) = \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}}{\rm{  =  40}}{\rm{.}}\frac{1}{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}}.\frac{1}{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}} = 40L\left[ {{e^{ - 2t}}.\sin t} \right].{L^{ - 1}}\left[ {{e^{ - 2t}}.\sin t} \right] \\ 
  =  > g(t) = {e^{ - 2t}}.\sin t*{e^{ - 2t}}.\sin t = \int_0^t {{e^{ - 2(t - u)}}.\sin (t - u).{e^{ - 2u}}.\sin u.du}  \\ 
  = {e^{ - 2t}}\int_0^t {(\sin t.\cos u - \cos t.\sin u)\sin u.du}  \\ 
  =  > g(t) = {e^{ - 2t}}\left[ {\sin t\underbrace {\int_0^t {\cos u.\sin u.du} }_{{I_1}} - \cos t\underbrace {\int_0^t {{{\sin }^2}udu} }_{{I_2}}} \right] \\ 
  =  > f(t) = 2\delta (t) + g(t) \\ 
 \end{array}$
You can evalue the integrals I1 and I2 yourself
