Originally Posted by

**AllanCuz** Hmmm...I believe you would have to use a power series representation here.

For $\displaystyle 0 \le x \le 1 $

$\displaystyle \frac{ ln(1-x) }{ x} = \frac{ -(x + \frac{x^2}{2} + \frac{x^3}{3}+...) }{x} = -(1+ \frac{x}{2} + \frac{x^2}{3}+...)$

If we take the integral of this,

$\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \int_0^1 (1+ \frac{x}{2} + \frac{x^2}{3}+...) = (1 + \frac{1}{2^2} + \frac{1}{3^2}+...) $

Note how this is the series,

$\displaystyle \sum \frac{1}{n^2} = \frac{ ( \pi)^2 }{6} $

So $\displaystyle -\int^{1}_{0} \frac{\ln(1-x)}{x} \ dx = \frac{ ( \pi)^2 }{6} $

Edit- Can you explain your way? I don't quite comprehend what you were trying to do but it seems like a cool exercise (so if you could, could you post your solution?)