Your idea is correct but the calculation is not. For $N$ moles of an ideal gas you have $U = cNRT$, look that if $T$ doesn't change, $U$ also doesn't change. So you start at the state $(T_1,V_1)$ and want to go to the state $(T_2,V_2)$. You then use two process corresponding to the following sequence of states:
$$(T_1,V_1)\to (T_2,V_1)\to (T_2,V_2)$$
On the first, you indeed have $\Delta U_1 = cNRT_2 - cNRT_1 = cNR(T_2-T_1)$ but the second one however has temperature equal to $T_2$ so that $\Delta U_2 = 0$. This implies that the total change in energy is $\Delta U = cNR(T_2-T_1)$ which is just the first term you wrote.
Edit: About your explanation to the second term, the quantity $Nk_B T/V$ is indeed pressure, since $PV = Nk_B T$. Now, because of that your integral gives in truth the negative of the amount of work on the second process. Recall that $\delta W = -PdV$ and so
$$W = -\int_\Gamma PdV$$
But by the first law of Thermodynamics, $\Delta U = W + Q$, so that change of energy is not made up just of work, but of heat also. By the formula for energy you could then see that for this process $Q = -W$.