Trying to figure out the Close Loop System transfer function of an op-amp, however the math isn't working out as it should.
Circuit In question:
simulate this circuit – Schematic created using CircuitLab
Using this as the foundation of calculating the Close loop transfer function: 
Where A is the Transfer function (open loop) of the LM1875 which taken from the bode plot obtained here: LM1875 DataSheet. TF should be $$\ A = H(s) = \frac{31.62}{\mathrm{707*10}^{-9}*s+1}$$
Where B is the transfer function of the negative feedback $$\ B = G(s) = \frac{R2}{R1}+1 = \frac{1000k\Omega}{1000k\Omega}+1 = 2$$
The close loop Equation $$\ CL(s) = \frac{H(s)}{1+H(s)*G(s)} $$
$$\ CL(s) = \frac{\frac{31.62}{\mathrm{707*10}^{-9}*s+1}}{1+(\frac{31.62}{\mathrm{707*10}^{-9}*s+1})*(2)} $$
$$\ CL(s) = \frac{\frac{31.62}{\mathrm{707*10}^{-9}*s+1}}{1+(\frac{63.24}{\mathrm{707*10}^{-9}*s+1})} $$
$$\ CL(s) = \frac{\frac{31.62}{\mathrm{707*10}^{-9}*s+1}}{\frac{\mathrm{707*10}^{-9}*s+1}{\mathrm{707*10}^{-9}*s+1}+(\frac{63.24}{\mathrm{707*10}^{-9}*s+1})} $$
$$\ CL(s) = \frac{\frac{31.62}{\mathrm{707*10}^{-9}*s+1}}{\frac{\mathrm{707*10}^{-9}*s+1+63.24}{\mathrm{707*10}^{-9}*s+1}} $$
$$\ \require{cancel} CL(s) = \frac{31.62}{\cancel{\mathrm{707*10}^{-9}*s+1}} * \frac{\cancel{\mathrm{707*10}^{-9}*s+1 }}{\mathrm{707*10}^{-9}*s+1+63.24} $$
$$\ \require{cancel} CL(s) = \frac{31.62}{\mathrm{707*10}^{-9}*s+64.24} $$
Using FVT as $$\ S\xrightarrow{}0 $$ :
$$\ \require{cancel} CL(0) = \frac{31.62}{\mathrm{707*10}^{-9}*(0)+64.24} = \frac{31.62}{64.24} = 0.4922 = DC Gain$$
Where did I go wrong?
I know this is wrong as the DC gain should be around 2V/V
Best Answer
I believe your mistake is to assume $$\ B = G(s) = \frac{R2}{R1}+1 = \frac{1000k\Omega}{1000k\Omega}+1 = 2$$ B is actually 1/2, as it is the output voltage divided by 2, that is $$\ B = G(s) = \frac{R1}{R1+R2} $$ Where the ratio comes from the \$R1,R2\$ voltage divider.
With this value of B, you would obtain $$\ CL(0) = \frac{31.62}{16.81} = 1.85$$