I was doing my laplace transform revision and I came across this problem. How to decompose this partial fraction ?
$$ \frac{145s}{(s^2+4)(s^2+4s+13)} +\frac{9s+55}{s^2+4s+13}$$
How to write the partial decomposition form and I much appreciated if you can show me the full step. I really forgot the partial decomposition. The given answer is $$\frac{16}{s^2+4} + \frac{9s}{s^2+4} + \frac{3}{s^2+4s+13}$$
$s^2+4$ and $s^2+4s+13$ are irreducible quadratics, so you’re really just decomposing
$$\begin{align*} \frac{145s}{(s^2+4)(s^2+4s+13)}&=\frac{As+B}{s^2+4}+\frac{Cs+D}{s^2+4s+13}\\\\ &=\frac{(As+B)(s^2+4s+13)+(Cs+D)(s^2+4)}{(s^2+4)(s^2+4s+13)}\;. \end{align*}$$
Equate numerators:
$$\begin{align*} 145s&=(As+B)(s^2+4s+13)+(Cs+D)(s^2+4)\\ &=(A+C)s^3+(4A+B+D)s^2+(13A+4B+4C)s+(13B+4D)\;, \end{align*}$$
so
$$\left\{\begin{align*} &A+C=0\\ &4A+B+D=0\\ &13A+4B+4C=145\\ &13B+4D=0\;. \end{align*}\right.\tag{1}$$
Solving the system $(1)$ is mildly annoying but routine; the solution is $A=9,B=16,C=-9$, and $D=-52$. Thus,
$$\frac{145s}{(s^2+4)(s^2+4s+13)}=\frac{9s+16}{s^2+4}-\frac{9s+52}{s^2+4s+13}\;,$$
and
$$\begin{align*} \frac{145s}{(s^2+4)(s^2+4s+13)}+\frac{9s+55}{s^2+4s+13}&=\frac{9s+16}{s^2+4}+\frac3{s^2+4s+13}\\\\ &=\frac{9s}{s^2+4}+\frac{16}{s^2+4}+\frac3{s^2+4s+13}\;. \end{align*}$$
