Inverse Laplace Transform partial fraction $\frac{2s}{(s+1)^2 + 4}$

Question

Can anybody help me with the answer of this question?

Find the inverse Laplace transform of: $$\frac{2s}{(s+1)^2 + 4}$$

Answer 1

$\newcommand{\L}{\mathscr{L}}$Hint:

Note that $$\L^{-1}\left(\frac{2s}{(s+1)^2+4}\right) = 2 \L^{-1}\left(\frac{s+1}{(s+1)^2+4}-\frac{1}{(s+1)^2+4}\right).$$ Using linearity, it is thus enough to find the inverse Laplace Transform of $\frac{s+1}{(s+1)^2+4}$ and of $\frac{1}{(s+1)^2+4}$. To find these, use standard inverse Laplace transforms along with the fact that $$\color{blue}{\L^{-1}\left( F(s-a)\right) = e^{at}f(t)},$$ where $F$ is the Laplace transform of $f$ (with $a=-1$ in your problem).

Answer 2

You can write $$\frac{2s}{(s+1)^2 + 4} = 2\left(\frac{s-1}{(s+1)^2 + 4}+\frac{1}{2}\frac{2}{(s+1)^2 + 4}\right)$$ then follow a table of Laplace transforms to get $$2(e^{at}\cos(2t) +\frac{1}{2} e^{at}\sin(2t))$$ so to finally get $$e^{at}(2\cos(2t) +\sin(2t))$$

It's a good practice to try to get to the closest known form and see if adding what is missing can lead to a second known term