Integration of $\frac{\ln(1-x)}{x}$

Question

How can I find the indefinite integral which is $$\int \frac{\ln(1-x)}{x}\text{d}x$$

I tried to use substitution by assigning $$\ln(1-x)\text{d}x = \text{d}v $$ and $$\frac{1}{x}=u$$ but, it is meaningless but true, the only thing I came up from integration by part is that $$\int \frac{\ln(1-x)}{x^2}\text{d}x = foo $$ and that has no help for me to find the integration $$\int \frac{\ln(1-x)}{x}\text{d}x$$

Answer 1

This is a "well-known" special function: $$\int \dfrac{\ln(1-x)}{x} \; dx = - \text{dilog}(1-x) $$ It is (provably) not an elementary function. In particular, there is no closed-form expression for it in terms of the functions familiar to the typical calculus student.

Answer 2

This integral has no primitive. Indeed the result is a so well known Special function called Logarithm Integral:

$$\int\frac{\ln(1-x)}{x}\ \text{d}x = -\text{Li}_2(x)$$

More here

https://en.wikipedia.org/wiki/Logarithmic_integral_function

Answer 3

The primitive is not an elementary function. Since in a neighbourhood of the origin we have: $$ -\log(1-x) = \sum_{n\geq 1}\frac{x^n}{n} $$ it happens that: $$ \int_{0}^{x}\frac{\log(1-z)}{z}\,dz = -\sum_{n\geq 1}\frac{x^n}{n^2} = \color{red}{-\text{Li}_2(x)}$$ for any $x\in(-1,1)$.