Question on the sum $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$ $f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$.
$f'(x) = \displaystyle\sum_{n=1}^{\infty}x^{n-1} = 1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$
If $f'(x) = \displaystyle\frac{1}{1-x}$, then $f(x) = \displaystyle\int\frac{1}{1-x}dx$
But $\displaystyle\int \frac{1}{1-x}dx = -\ln(1-x) + C$, and so $f(x) = -\ln(1-x) + C$
Do we need a definite integral here so there will be no $C$ and will agree with how $f(x)$ was originally defined? If so what would the definite integral be? Or maybe there is no definite integral at all?

There are various things going on here. One is the issue of the $+C.$ What you wrote might better be put

If $f'(x) = \displaystyle\frac{1}{1-x}$, then by definition of the indefinite integral $\displaystyle\int\frac{1}{1-x}dx=f(x)+C$

Also , by definition of definite integrals
If $f'(x) = \displaystyle\frac{1}{1-x}$, on $[a,b]$ then $ \displaystyle\int_a^b\frac{1}{1-t}dt=f(b)-f(a)$ 
Note that an arbitrarily chosen constant of integration $C$ will cancel out.
I changed the variable of integration so that I could say that for $|x| \lt 1,$ $$\int_0^x\frac1{1-t}dt=-\ln(1-t)\mid_0^x=\left(-\ln(1-x)\right)-(-\ln(1-0))=\ln\left(\frac1{1-x}\right)-0$$ 
There is also the matter of term by term integration and differentiation of power series. Notice that this is one of the interesting cases where 
$f(x) = = x + \frac{x^2}{2} + \frac{x^3}{3} + ... = -\ln(1-x)$ for $|x| < 1$ and also $x=-1$ but not $x=1.$
$f'(x) =  1 + x + x^2 + x^3 +... = \frac{1}{1-x}$ for $|x| < 1$ but not for  $x= -1$ or $x=1.$
