Find $\int \frac{\ln(1-x)}{x}dx$

Question

I would like to find $$\int \frac{\ln(1-x)}{x}dx$$

On the site where I found this the problem was actually slightly different; the question was to evaluate $$\int_0^1 \frac{\ln(1-x)}{x}dx$$ which can be solved using its Maclaurin series expansion and the solution for the Basel problem, but I was wondering if there's a way to directly find the antiderivative of the integrand. I have tried substitutions included hyperbolic and polynomial ones, such as $x=\sinh^2 x$, but I haven't managed to get anywhere with it.

Would you mind giving me a push in the right direction?

Thank you for your help.

Answer 1

The Euler dilogarithm function $~Li_2(x)~$ defined for $0 ≤ x ≤ 1$ by $$Li_{2}(x)=\sum_{n\ge1}\dfrac{x^n}{n^2}=-\int_{0}^{x}\frac{\ln{(1-t)}}{t}dt$$is one of the lesser transcendental function. Nonetheless, it has many intriguing properties and has appeared in various branches of mathematics and physics.

So $$\int_0^1 \frac{\ln(1-x)}{x}dx=-Li_2(1)=-\dfrac{\pi^2}{6}$$

You may read the following links (and the references therein) for further knowledge about this type of integrals:
"Dilogarithm Identities" by Anatol N. Kirillov
Calculate a $\operatorname{Li}_{2}(-1)$ using Integral Representation
Dilogarithm by Wolfram MathWorld
The Dilogarithm Function by Don Zagier
Spence's function by Wikipedia