Partial decomposition of $\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$ Recently I have come arcross the following fraction

$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$$

Hence I have encountered this fraction within a task of integration I want to do a partial decomposition. First of all I rewrote it as following
$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}=\dfrac{Az+B}{2z^2+3}+\dfrac{Cz+D}{(s^2+1)z^2+1} $$
which yields to a system of equations for $A,B,C$ and $D$
$$\begin{align}
0&=A(1+s^2)+3C\\
0&=B(1+s^2)+3D\\
0&=s^2A+2C\\
1&=s^2B+2D
\end{align}$$
Solving this system gives $A=C=0$ and $B=\dfrac{3}{s^2-2}$ and $D=-\dfrac{s^2-1}{s^2-2}$. By plugging in these values we arrive at

$$\dfrac{3}{(s^2-2)(2z^2+3)}-\dfrac{s^2-1}{(s^2-2)((s^2+1)z^2+1)}=\dfrac{s^2(z^2-3)+5z^2+6}{(s^2-2)(2z^2+3)((s^2+1)z^2+1)}$$

which is in fact not the wanted fraction. From the original task(solution to Problem 2 by ysharifi) I am going through I know, that the right partial decomposition should look like 

$$\dfrac1{3s^2+1}\left(\frac3{2z^2+3}-\frac1{(s^2+1)z^2+1}\right)$$

but honestly speaking I have no clue how to get to this. Could someone please go through the whole process with me. I am quite confused right now.
Thanks in advance.

Even though we could solve this using the method of residues, I'll follow your method
\begin{equation}
 \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
 =
 \frac{Az + B}{2z^2 + 3}
 +
 \frac{Cz + D}{(s^2+1)z^2 + 1}
 \tag{1}
\end{equation}
which is
\begin{equation}
 \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
 =
 \frac{(Az + B)((s^2+1)z^2 + 1) + (Cz + D)(2z^2 + 3)}{(2z^2 + 3)((s^2+1)z^2+1)}
\end{equation}
Expanding
\begin{equation}
 \frac{ B+3D + (A+3C)z + (B+2D+Bs^2)z^2 + (A+2C+As^2)z^3}{(2z^2 + 3)((s^2+1)z^2+1)} 
\end{equation}
By identification of coefficients, we have
\begin{align}
 B+3D &= 0\\
 A+3C &= 0 \\
 B+2D+Bs^2 &= 1 \\
 A+2C+As^2 &= 0
\end{align}
From the first two, we get $A = -3C$ and $B= - 3D$, so replace in the last two,
\begin{align}
 -D -3s^2D &= 1\\
 -C -3s^2C &= 0
\end{align}
which means that $C =0$ (also $A=0$) and
\begin{equation}
 D = -\frac{1}{1+3s^2}
\end{equation}
So replacing in $(1)$, we get
\begin{equation}
 \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
 =
 \frac{1}{1+3s^2}
 \Bigg(
 \frac{ 3}{2z^2 + 3}
 -
 \frac{ 1}{(s^2+1)z^2 + 1}
 \Bigg)
\end{equation}
