Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1
Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1<x<1$ using the power series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$
I did this using term by term integration
$\int \frac{1}{1-x} dx = (-1) \ln (1-x) + C = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$
LHS: $\ln (1) - \ln (1-x) = \ln \frac{1}{1-x} + C$
What do I have to do with the right hand side of the equation? And afterwards, what is the reason for substituting $x=0$ to find the value of $C$ ? Where did the $x=0$ came from?

You are on the right track. For $ |x|<1$ let
$$f(x)= \sum_{n=1}^{\infty}\frac{x^n}{n}.$$
Then
$$f'(x)= \sum_{n=1}^{\infty}x^{n-1}= \sum_{n=0}^{\infty}x^{n}= \frac{1}{1-x}.$$
Hence
$$f(x)= - \ln (1-x)+c.$$
With $x=0$ we see that $c=0.$
Therefore
$$f(x)= - \ln (1-x)= \ln (1)-\ln (1-x)= \ln ( \frac{1}{1-x}).$$
