I was requested to determine the power representation $$\int \frac{\ln(1-x)}{x}dx$$ in powers of $x$. My approach was rather simple.
$i)$ Let $f(x)=\frac{\ln(1-x)}{x}=\frac{1}{x}\ln(1-x)$. It is not hard to show that
$$\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}{n}$$
and therefore
$$\frac{1}{x}\ln(1-x)=-\sum_{n=1}^\infty \frac{x^{n-1}}{n}$$
$ii$) We now have $$\int \frac{\ln(1-x)}{x}dx=\int-\sum_{n=1}^\infty \frac{x^{n-1}}{n}dx=-\sum_{n=1}^\infty \int\frac{x^{n-1}}{n}dx=-\sum_{n=1}^\infty \frac{x^{n}}{n^2}+C$$
$iii)$ ...
All there's left to do is to find the value of $C$ (point $iii$), which is where I struggled, for I can not let $x=0$ nor $x=1$ (results in undefind expression on the integral), and other values of $x$ have not prove to be useful to me. Is my procedure correct and, if so, how can I find the value of $C$?