Different functions having the same derivative

Question

Let $g(x) = \frac{1}{1-x}$ and $h(x) = \frac{x}{1-x}$:

Their derivatives are:

$$\frac{d}{dx}g(x)=\frac{1}{\left(1-x\right)^2}$$

$$\frac{d}{dx}h(x)=\frac{1}{\left(1-x\right)^2}$$

Integrating the RHS I should get back the initial function(s):

$$\int \frac{1}{\left(1-x\right)^2}\;\mathrm dx=\frac{1}{1-x} + C$$

But I obviously get only one of the expressions, namely $g(x)$. So my question is how can both functions have the same derivative even though when integrating the derivative, I get back only one of the expressions?

What am I missing? Does it have something to do with the $C$ constant?

Answer 1

Note that $$\frac1{1-x}-1=\frac{x}{1-x}$$

Answer 2

The functions $1$ and $0$ are also different functions with the same derivative. They differ by a constant, just like your functions $g$ and $h$.

Answer 3

$g(x)=\dfrac{1-x+x}{1-x}=1+h(x)$.

So, $g'(x)=h'(x)$.