How can this function have two different antiderivatives?

It is not really a contradiction, since difference of the two functions is constant: $$ \frac1{1-u(t)} - \frac{u(t)}{1-u(t)} = \frac{1-u(t)}{1-u(t)}=1. $$ (Derivative of a constant function is zero. Primitive function is determined uniquely up to a constant.)

I have not verified that the derivatives are correct, but notice that $$\frac1{1-u(t)}-\frac{u(t)}{1-u(t)}=1$$

that is, the difference between both antiderivatives is constant. This implies that the derivatives of both functions are the same.

$$\frac{d}{dt} \frac{u(t)}{1-u(t)}=\frac{d}{dt} \left(\frac{1}{1-u(t)}-1\right)=\frac{d}{dt} \frac{1}{1-u(t)}$$

Note that $1$ is just a constant so it vanishes. But when computing the antiderivative you will specify the constant according to initial conditions.