Is this the true expression for isochoric enthalpy of non-ideal gases?
The derivation for the first part of your question is shown below. I'm not sure what you are trying to do in the second part. 

The heat capacity is the heat required to raise the temperature of  1 mole of substance by 1 K. Suppose  for the moment that heat capacity C is constant over a temperature range $T_1 $ to $T_2$, then the heat q required to raise 1 mole of substance  over this temperature range is 
$$q=C(T_2-T_1) = C\Delta T$$
If the heat capacity is not constant, which is generally the case as real molecules have vibrational and rotational energy levels then, 
$$ q=\int _{T_1} ^{T_2}C(T)dT$$

where in many cases the heat capacity can be approximated as $C(T)=a+bT+cT^{-2}$. For a gas the partition function can be calculated, if spectroscopic values are known, and $C(T)$ calculated that way.

The heat absorbed depends on the path taken to go between the two end points so that C is not well defined. To make it well defined either T or P must be constant. 

At constant volume, $q_V=C_V\Delta T$ and the work done as expansion against external pressure is zero, thus from the first law with internal energy as U,
$$ \Delta U = C_V\Delta T $$

and a similar integral as described above is used when the temperature range is too large for $C_V$ to be considered a constant. (This equation shows also, in the limit of small changes,  that heat capacity is the slope of internal energy vs. temperature.)

In the case of constant pressure $q_P=C_P\Delta T$. The change in enthalpy is 
$$\Delta H = C_P\Delta T$$
and so in general
$$ \Delta H=\int _{T_1} ^{T_2}C_P(T)dT$$

To demonstrate that $q_P=\Delta H$ we note that in this case the volume is not constant and so the heat transferred at constant pressure $q_P$  is related to internal energy change $\Delta U$ as 
$$ \Delta U = q_P-P\Delta V$$
where $P\Delta V$ is the work done. Rearranging and expressing the $\Delta$'s directly gives
$$ q_P= (U_2+PV_2) - (U_1 +PV_1)$$
from which using the definition of enthalpy $H=U+PV$ gives $q_P=H_2-H_1$ and shows that $q_P=\Delta H$.