The internal energy is clearly a state function, and thus its differential must be an exact differential. $$dU = \text{ ?}$$ $$= đQ - đW$$ $$ = đQ - pdV \text{ only when change is quasistatic}$$

This $−pdV$ term can be a bit confusing at first. You are accustomed to work being $Fdx$. With a little thought, you can recognize pp as the force per unit volume, and the ratio of $dV$ and $dx$ as the area. The minus sign comes from the fact that a positive pressure pushes outwards. What is this $đQ$? As it turns out, we can define a state function $S$ called entropy and so long as a process is done reversibly $$đQ = TdS \text{ only when change is quasistatic}$$ so we find out that $$dU = TdS - pdV$$

The fact that the $T$ in this equation is actually the physical temperature measured by our thermometers was originally an observation based on experiment. At this point, the entropy $S$ is just some weird heat-related state function.

If you decide to get a thermodynamics tattoo, my recommendation would be to choose the thermodynamic identity

$$dU = TdS - pdV$$

It is far and away the most fundamental and essential equation, and one which you will need to come back to again and again. It contains hidden within it (if you remember the First Law) the thermodynamic definition of entropy.