Gravitational collapse (conservation of energy) During gravitational collapse, gravitational potential energy of the gas is converted to its internal kinetic energy so the internal energy of the clous of gas is said to be increased
But isn't gravitational potential energy included in the internal energy? Shouldn't the internal energy remains constant overall?
 A: 
During gravitational collapse, gravitational potential energy of the gas is converted to its internal kinetic energy so the internal energy of the cloud of gas is said to be increased.

Like CuriousOne said, this depends on how you define the internal energy of the cloud. IMHO you can understand this better by thinking about a single falling body: a brick. 

But isn't gravitational potential energy included in the internal energy?

Yes. Gravitational potential energy is internal energy. When you throw a brick up into the air, you do work on it, you add energy to it*. You give it kinetic energy which is converted into gravitational potential energy by gravity. As a result the mass of the brick is increased, because what was external kinetic energy has become internal kinetic energy. When you then drop the brick, some of this internal kinetic energy is converted back into external kinetic energy. If you dissipate this by catching the brick and letting the kinetic energy radiate away, you're back where you started, and the mass of the brick is reduced to what it it was. Check out the mass deficit.  

Shouldn't the internal energy remains constant overall?

The energy remains constant overall. If your gas cloud was surrounded by a reflective balloon skin which prevented energy radiating away, then the internal energy would remain constant. But you don't, so it doesn't. 
Note that gravity doesn't perform work on a falling body in that it doesn't add energy to it. It merely converts internal kinetic energy into external kinetic energy. You do work on the body when you throw it up into the air. 
* Conservation of p=mv momentum tells us there is an effect on the Earth, but it  doesn't move in any detectable fashion, and the KE=½mv² kinetic energy is not shared equally. So we ignore the effect upon the Earth. 
