Inverse Laplace Transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. I am trying to figure out the inverse laplace transform of $\dfrac{6s -19}{s^2 - 6s + 13}$. Looking at my table of Laplace Transforms in my textbook, it seems that either I must break up this fraction using partial fractions into linear terms or I need to use some trick that I'm missing. I know that the denominator cannot be broken up further into linear terms, since if I try to solve $s^2 - 6s + 13$ I get complex roots, so I'm unsure what to do. Any help is appreciated.

Given: 
$$Y(s)=\frac{6s-19}{s^2-6s+13}$$
APPROACH #1:
Completing the square of the denominator gives
$$Y(s)=\frac{6s-19}{(s-3)^2+4}=\frac{6s}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}$$
Algebraic manipulation of each term gives an equivalent equation in a more appropriate form (reasons behind this step will become apparent shortly)
$$Y(s)=6\frac{s-3+3}{(s-3)^2+4}-\frac{19}{(s-3)^2+4}=6\frac{s-3}{(s-3)^2+4}-\frac{1}{(s-3)^2+4}$$
or
$$Y(s)=6\frac{s-3}{(s-3)^2+4}-\frac{1}{2}\frac{2}{(s-3)^2+2^2}$$

From the transform tables, we use the pairs:
  
  
*
  
*$e^{at}cos(bt)\leftrightarrow \frac{s-a}{(s-a)^2+b^2}$
  
*$e^{at}sin(bt)\leftrightarrow \frac{b}{(s-a)^2+b^2}$
  

Identifying: $a=3$ and $b=2$, we get
$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} =  6e^{3t}cos(2t)-\frac{1}{2}e^{3t}sin(2t)$$

APPROACH #2: 
Invoking the idea of partial fraction decomposition, we write
$$\frac{6s-19}{s^2-6s+13}=\frac{A}{s-3-2i}+\frac{B}{s-3+2i}$$
or after multiplying by $s^2-6s+13$
$$(s-3+2i)A+(s-3-2i)B=6s-19$$


*

*If $s=3-2i$:
$$0+(3-2i-3-2i)B=6(3-2i)-19\Rightarrow B=3-\frac{1}{4}i$$

*If $s=3+2i$:
$$(3+2i-3+2i)A+0=6(3+2i)-19\Rightarrow A=3+\frac{1}{4}i$$
Therefore
$$Y(s)=\frac{3+\frac{1}{4}i}{s-3-2i}+\frac{3-\frac{1}{4}i}{s-3+2i}$$

From the transform tables, we use the pair:
  
  
*
  
*$\alpha e^{(a+jb)t}\leftrightarrow \frac{\alpha}{s-(a+jb)}$
  
  
  to invert the terms.

Identifying: $a=3$ and $b=2$, we get
$$ y(t) = \mathcal{L^{-1}} \left \{ Y(s) \right \} =  (3+\frac{1}{4}i)e^{\left (3+2i  \right )t}+(3-\frac{1}{4}i)e^{\left (3-2i  \right )t}$$
To show that this is equivalent to the one found above, we use little complex algebra:
\begin{align}
y(t) &= (3+\frac{1}{4}i)e^{\left (3+2i  \right )t}+(3-\frac{1}{4}i)e^{\left (3-2i  \right )t} \\
 & = e^{3t}\left [ (3+\frac{1}{4}i)e^{2it}+(3-\frac{1}{4}i)e^{-2it} \right ] \\ 
 & = e^{3t}\left [ 3(e^{2it}+e^{-2it})+\frac{1}{4}i(e^{2it}-e^{-2it})\right ]\\
 & = e^{3t}\left [ 3(2cos(2t))+\frac{1}{4}i(2isin(2t))\right ] \\ 
 & = e^{3t}\left [ 6cost(2t)-\frac{1}{2}sin(2t)\right ] \\
& = 6\;e^{3t}cost(2t)-\frac{1}{2}\;e^{3t}sin(2t)
\end{align}
