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\( \begin{align} H&=\frac{j\omega+8}{(j\omega)^2+4*j\omega+13}\\ H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2} \end{align} \)

From the \(s\) in the numerator, you can see that \(A\) is 1. That means the numerator so far is \(1(s+2)\) Therefore, there is already a constant 2 up top. To get the total of 8, then \(B\) needs to be 2. That is:

\( \begin{align} H&=\frac{s+8}{(s+2)^2+(3)^2}=\frac{A(s+2)+B(3)}{(s+2)^2+(3)^2}=\frac{1(s+2)+2(3)}{(s+2)^2+(3)^2} \end{align} \)

meaning

\( h(t):=e^{-2t}\left(1*\cos(3t)+2*\sin(3t)\right)~u(t) \)

It is now possible to do inline math such as $$c=\sqrt{a^2+b^2}$$ with double dollar signs around the math.