How can this function have two different antiderivatives?

I'm currently operating with the following integral:

$$\int\frac{u'(t)}{(1-u(t))^2} dt$$

But I notice that

$$\frac{d}{dt} \frac{u(t)}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$

and

$$\frac{d}{dt} \frac{1}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$

It seems that both solutions are possible, but that seems to contradict the uniqueness of Riemann's Integral.

So the questions are:

1. Which one of them is the correct integral?
2. If both are correct, why the solution is not unique?
3. The pole at $u(t)=1$ has something to say?

Solution 1:

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.)

Solution 2:

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.

Solution 3:

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

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