How can objects with no net acceleration gain energy? Suppose I have a box on the floor. I lift this box up, being very careful to make sure it rises with a constant upward velocity. (Yes, I will need to accelerate it to start, but assume I do that quickly.) Now, as this box rises with this constant velocity, it gains potential energy, but how? From my understanding, the work done on a body is the integral of its net force over the path it takes. If velocity is constant there is no net force, so there would be no net work done, so somehow this potential energy was created without any work being done ... this makes no sense.
I know this must be wrong since energy is always conserved, but I can't seem to find a flaw in this reasoning.
 A: Your reasoning is completely right. The kinetic energy of the box is $mv^2/2$, so if $v$ is constant, the energy doesn't change. You do positive work $W$ on the box by pushing it up, while gravity does negative work $-W$ by pulling it down, so there's zero net work. No contradiction!
The reason you might be confused is, you might be wondering, doesn't the gravitational potential energy go up? Well, there's a subtlety here about using potential energy. Potential energy is a property of entire systems, in this case of the Earth and box. If you were considering the system of the Earth and box, you would say that you do work on a part of the system (the box), and the potential energy of the system goes up. You shouldn't count the gravitational force on the box in this case, because it's internal to the system. 
On the other hand, if you consider the system of the box alone (as I did in the first paragraph), then there are two external forces: your force, and that of gravity. And there's only one contribution to the energy, the kinetic energy of the box itself. So in both cases, conservation of energy works. 
Basically: you can count gravity as an external force that can do work on your system, or an internal force that is associated with a potential energy. But you can't do both at once -- that will double count it!
A: It’s true that there is no net work done. But this just implies that there is no change in kinetic energy, which is true since velocity is constant. The main point here is that the force that you apply is a non-conservative force and work done by non conservative force is change in total energy. Since this work is positive, the total energy is increasing and at the same time, this keeps on “getting stored” (I am using hand wavy terms to give intuition) since the negative of the work done by the weight is change in potential energy. To summarise, here are the expressions that lead to the above conclusion; you may refer any standard textbook to see how these are derived (Of the top of my head, Ch. 1, Sec. 1.3 of ‘Fundamentals of Vibrations’ by Leonard Meirovitch contains this):
$$\int_{r_1}^{r_2}\mathbf{F_{nc}}.d\mathbf{r}=E_2-E_1$$
where $\mathbf{F_{nc}}$ is the net non-conservative force and $E$ is the total energy. 
$$\int_{r_1}^{r_2}\mathbf{F_c}.d\mathbf{r}=V_1-V_2$$ 
where $\mathbf{F_c}$ is the net conservative force and $V$ is the potential energy.
$$\int_{r_1}^{r_2}\mathbf{F}.d\mathbf{r}=KE_2-KE_1$$ 
where $\mathbf{F}$ is the net force and $KE$ is the kinetic energy.
