I had a test yesterday and couldn't figure out the answer to this question. Was wondering if someone could walk me through the solution. $$\dfrac{(13s+3)}{(s^2+2s+5)}$$
I figured I had to complete the square in the denominator then split the fraction up into $$\dfrac{13s}{(s+1)^2+4} + \dfrac3{(s+1)^2+4}$$
That would allow me to solve the second fraction as $3/2 \exp(-t) \sin(2t)$.
I couldn't figure how to solve for the first fraction unless I shouldn't have split them up at all.
So what is the proper way to solve it?
Since you noticed that you needed to complete the square in the denominator, let that dictate the form of the new, preferred numerator as follows: $$ {13s+3\over s^2+2s+5}={13s+3\over (s+1)^2+2^2}={A\color{blue}{(s+1)}+B\cdot \color{red}{2}\over \color{blue}{(s+1)}^2+\color{red}{2}^2}, $$ where equating coefficients between the first and last versions of the numerator yields $A=13$, $B=-5$.
This last form is quite helpful since a standard table of Laplace transforms shows \begin{align} \mathscr{L}\left\{e^{-at}\cos(bt)\right\}={s+a\over (s+a)^2+b^2},\\ \mathscr{L}\left\{e^{-at}\sin(bt)\right\}={b\over (s+a)^2+b^2}, \end{align} so \begin{align} \mathscr{L}^{-1}\left\{{13s+3\over s^2+2s+5}\right\}&=\mathscr{L}^{-1}\left\{{A(s+1)+B\cdot 2\over (s+1)^2+2^2}\right\}\\ &=A\,\mathscr{L}^{-1}\left\{{s+1\over (s+1)^2+2^2}\right\}+B\,\mathscr{L}^{-1}\left\{{2\over (s+1)^2+2^2}\right\}\\ &=Ae^{-t}\cos(2t)+Be^{-t}\sin(2t),\\ &=13e^{-t}\cos(2t)-5e^{-t}\sin(2t). \end{align}