Does the internal energy of single-phase materials depend primarily on temperature or does it depend on pressure as well? I am confused about the relation between the internal energy and temperature of ideal gases and single-phase materials.
The following two statements from two distinct sources are mainly the reasons:

In single-phase, materials such as solids, liquids, and gases, the
  internal energy of a system depends primarily on the temperature.

quoted from Introduction to Thermal and Fluids Engineering by Deborah A. Kaminski and Micheal K. Jensen

... in an isothermal process, the internal energy of an
  ideal gas is constant. ...
  Note that this is true only
  for ideal gases; the internal energy depends on pressure as well as on
  temperature for liquids, solids, and real gases.

quoted from  Isothermal Process, Wikipedia
Does the second article not suggest that they are single-phase when referring to liquids, solids and real gases? And also, why does the internal energy of single-phase materials depend primarily on temperature?
 A: The 1st law of thermodynamics states that the change of the internal energy is $dU=\delta Q + \delta W$. that is the sum of the absorbed heat and work, so if you want to change the internal energy you have to transfer either heat or work. When combined with the 2nd law you may write this as the Gibbs equation spelling out the various interactions explicitly: $dU = TdS -pdV + \mu dN - edE - mdH + etc.,$
For and ideal gas we assume only thermal and mechanical interaction, ie., $dU=TdS-pdV$, and assume that $U=f(T)$. This is essentially a definition but derived from experience with dilute gasses irrespective of the interaction it participates in. This has nothing to do with the process, instead it is in the nature of the material system, ie., the ideal gas. 
Practical liquids and solids are essentially incompressible with everyday pressures, $dV=0$, therefore the mechanical interaction term is missing from the Gibbs' equation, and what remains is the thermal interaction and everything else, if any: $dU = TdS + \mu dN - edE - mdH + etc.,$ But if the material is electrically/magnetically neutral and the process is such that $dN=0$, then all you have left is $dU=TdS$.
A: For a single phase closed system (solid, liquid, or gas), the differential change in internal energy dU as a function of dT and dV is given by the following general equation (neglecting electric and magnetic fields):
$$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$
Are you familiar with this equation?  For an ideal gas, the term in brackets is zero, and for an incompressible solid or liquid, dV is zero. So, for such ideal materials, dU is a function only of dT. And many solids, liquids, and gases approach these ideal situations.
Can you write out the expression for dV in terms of dP and dT?  When you combine this with the above equations, you get the relationship for dU in terms of dT and dP.
