I'll try to get at the result from first principles, let's start with $U = U(S,V,N)$. I'll just assume constant mole number, so I'll write $U(S,V)$.
We get the enthalpy by taking the "Legendre transform" of the internal energy.
$$H \equiv U[p]$$
which gives, $H = U + pV$ which is function of $S$ and $p$, i.e $H = H(S,p)$
taking the differential
$$\mathrm{d}H = \mathrm{d}U + p\mathrm{d}V + V\mathrm{d}p$$
$$\mathrm{d}U = \overbrace{\left(\frac{\partial U}{\partial S}\right)_V}^T\, \mathrm{d}S + \overbrace{\left(\frac{\partial U}{\partial V}\right)_S}^{-p}\, \mathrm{d}V$$
$$\mathrm{d}U = T\, \mathrm{d}S -p\, \mathrm{d}V$$
Using this,
$$\mathrm{d}H = T\, \mathrm{d}S + V\,\mathrm{d}p \tag{*}$$
imposing constant pressure conditions,
$$\mathrm{d}H = T\, \mathrm{d}S = \delta q_p $$
Hmm, now we that we have identified enthalpy as the heat exchanged at constant pressure.
Now, $q_p = n\int_{T_i}^{T_f}C_{p,m}\mathrm{d}T$ and finally,
$$\Delta H = \int_{i}^{f}\mathrm{d}H = q_p = n\int_{T_i}^{T_f}C_{p,m}\mathrm{d}T $$
Your last equation seems fishy, and I don't see where it comes from but the other two seem fine.
EDIT
So after my discussion with the OP in the comments section I was informed that we are not allowed to impose constant pressure, which I assume implicitly based on the equation he quoted from his lecture notes. If that is indeed the case, then the equation provided in his notes is incomplete. I just looked at @ChesterMiller's answer and that confirms it too.
So basically, how I would proceed is start with $(*)$
Now, $H = (S,p)$, but $S = S(T,p)$ too, so let's derive and invoke the "2nd $T\mathrm{d}S$ equation.
$$\mathrm{d}S(T,p) = \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T + \left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p
$$
multipling the LHS and RHS by $T$
$$T\,\mathrm{d}S = T \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T + T \left(\frac{\partial S}{\partial p}\right)_T \mathrm{d}p
$$
Using a Maxwell relation
$$T\,\mathrm{d}S = T \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T - T \left(\frac{\partial V}{\partial T}\right)_p \mathrm{d}p
$$
Almost done,
$$T\,\mathrm{d}S = C_p \mathrm{d}T - T V \alpha \mathrm{d}p $$
Putting this in $(*)$
$$\mathrm{d}H = C_p \mathrm{d}T - T V \alpha \mathrm{d}p + V \mathrm{d}p$$
or $$\mathrm{d}H = C_p \mathrm{d}T - V[1-T \alpha] \mathrm{d}p $$
which is the equation @ChesterMiller quoted. Now if you impose constant pressure, then you recover the relation I originally derived and it resembles the one you originally quoted. If you don't wish to do that, then this is the complete expression. To get the relation at constant volume, proceed as @ChesterMiller did in his answer.