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:


*

*Which one of them is the correct integral?

*If both are correct, why the solution is not unique?

*The pole at $u(t)=1$ has something to say?


I will explain the concept using the derivative since you are already pretty familiar with that. Lets define to $\int \:f\left(x\right)dx$ to be some fucntion where $\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)$. (Note this is not the exact definition of an anti-derivative, but an intuitive way of thinking about it.)
So if $g\left(x\right)=x^2,\:\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)=2x$ And in reverse,$\int \:f\left(x\right)dx=x^2$
But what happens if $g\left(x\right)=x^2+5,\:\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)=2x$, Notice the 5 is a constant and disappears when taking the derivative. If we apply the reverse, we have no way of getting back the 5. In reverse you still get $\int \:f\left(x\right)dx=x^2$. 
So now here comes the problem, when you are going in reverse you have no idea what the constant is. This is purely because the constant disappears when taking the derivative. For example $\frac{d}{dx}\left(x^2+8\right)=\frac{d}{dx}\left(x^2+5\right)=\frac{d}{dx}\left(x^2+1\right)$! So when finding the Anti-Derivative you will get a function plus or minus some constant.
