When can we separate the terms of a line integral? In one of my thermodynamics lectures, I came across something of the from
$S = \int \frac{dU+pdV}{T}$ which I know to be a line integral in differential form. I saw that in a problem this was simplified to $S= \int\frac{dU}{T}+\int\frac{pdV}{T}$. I was under the impression that we cannot simply separate the terms like this for a line integral. Is this a consequence of the fact that $S$(entropy) is a proper differential?
 A: $S$ is a function of state, so as long as we move from state $A=(U_0,V_0)$ to state $B=(U_1,V_1)$ in a reversible way, the change in entropy $\Delta S = S(B) - S(A)$ is independent of the path taken. So we can choose to go from $A$ to $B$ via $C=(U_1,V_0)$. As we go from $A$ to $C$ we keep $V=V_0$ constant, and as we go from $C$ to $B$ we keep $U=U_1$ constant. Then we can see that
$\displaystyle \Delta S = \int_A^B \frac{dU+p \space dV}{T} = \int_A^C \frac{dU+p\space dV}{T} +\int_C^B \frac{dU+p\space dV}{T} = \int_A^C \frac{dU}{T} +\int_C^B \frac{p\space dV}{T}$
A: If you wanna be technical about it, you can think of saying that U is some function of temperature, and the $ \frac{P}{T}$ ratio is some function of volume.
For an ideal gas, We can use the ideal gas law to turn the function which we are integrating into one of volume.
$$ \frac{P}{T} = \frac{nR}{V}$$
However, if you think of it from a purely mathematical standpoint, we are actually turning $dq$ into an exact differential via an integrating factor to arrive at a state function. Putting more precisely, if $S(U,V)$ then(*):
$$ (\frac{\partial S}{\partial U})_V = \frac{1}{T}$$
And,
$$ (\frac{\partial S}{\partial V})_U = \frac{P}{T}$$
The existence due to this condition(something about...curl?):
$$ \frac{ \partial }{\partial V} (\frac{\partial S}{\partial U})_U = \frac{ \partial}{\partial T}(\frac{\partial S}{\partial U})_V $$
Working out lhs:
$$ \big[ \frac{ \partial }{\partial V} (\frac{1}{T})\big]_U = \big[\frac{ \partial }{\partial V} (\frac{nC_v}{U}) \big]_U =0$$
And, RHS:
$$ \big[ \frac{ \partial}{\partial T}(\frac{\partial S}{\partial U})_V \big]=  \big[ \frac{ \partial}{\partial T} \frac{P}{T} \big]_V =  \big[\frac{ \partial}{\partial T} \frac{nR}{V} \big]_V = 0$$
Hence $dS$ is an exact differential and we must be able to find a state function of it.
Edit: A point to note maybe that when you have a process, the variables all change together that is a state maybe specified by the variables of $(P,V,T)$ so you can think of the integration as going from a state with state variables $(P,V,T)$ to one with say something like $(P',V',T')$

*: Those variables let us equate the partials when we take differential
For proving that $ \frac{1}{T}$ is integrating factor: See this question I asked here
Might want to see gradient theorem and independence of path conditions wiki
About the point on curl, I had made a stack post about in math stack exchange here
