<s>[INST] 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?
 [/INST] 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 $x=1.$
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