Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature? In all of my physical chemistry books I find the same expression for estimating the equilibrium constant of a reaction at a non-standard temperature. The derivation starts with $R \ln K_\text{eq} = -\Delta g^\circ/T = -(\Delta h^\circ/T - \Delta s^\circ)$. Then one takes the temperature derivative of each side assuming that $\Delta h^\circ$ and $\Delta s^\circ$ can both be considered constants, which eliminates the entropy term. The equation 
$\mathrm{d}\ln K = (1/R)(\Delta h^\circ/T^2)\mathrm{d}T$ is then integrated between $T_1$ and $T_2$ to give
$\ln(K_2/K_1) = (\Delta h^\circ/R)(1/T_1 - 1/T_2)$.

Here's the question: $\Delta h_\text{r}^\circ$ is really NOT a constant, but a function of temperature. Assuming constant heat capacities, $\Delta h_\text{r}^\circ$ at $T = \Delta h_\text{r}^\circ + \Delta c_p(T - T_0)$. Likewise $\Delta s_\text{r}^\circ$ at $T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$. Where $\Delta c_p$ is the difference in the heat capacity of the products and the reactants.

I have tried to insert these expressions into the typical derivation but am getting wonky results. Can someone help me with the derivation or point me to a source where $\ln K$ vs. $T$ is derived with temperature-sensitive reaction enthalpies and entropies.  

Edit 4/19/2015: Would it be more effective to invoke $\mathrm{d}\Delta g = \Delta v\mathrm{d}P - \Delta s\mathrm{d}T$ which at constant pressure, and in the case of free energy of reaction, gives $\mathrm{d}\Delta g_\text{r}^\circ/\mathrm{d}T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$ ?

Thanks in advance for any help.
For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$.  But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$

So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$