This looks awfully familiar

$\displaystyle{I=\int _{ 0 }^{ 1 }{ \cfrac { \ln { (1-x) } }{ x } dx } \\ I=\int _{ 0 }^{ 1 }{ (-1-\cfrac { x }{ 2 } -\cfrac { { x }^{ 2 } }{ 3 } ............)dx } \\ I=-(\cfrac { 1 }{ 1.1 } +\cfrac { 1 }{ 2.2 } +\cfrac { 1 }{ 3.3 } +\cfrac { 1 }{ 4.4 } ...........)\\ I=-(\sum _{ r=1 }^{ \infty }{ \cfrac { 1 }{ { r }^{ 2 } } } )=-\cfrac { { \pi }^{ 2 } }{ 6 } }$ I don't Know proof of this summation , Since I don't know about gamma,beta,zeta functions. But As Ronak is already proved beautifully this series some day's ago .

Check out here Ronak agarwal's proof

Using the taylor series: $\displaystyle{\int_{0}^{1}{ \frac{ln(1-x)}{x} } dx} = -\displaystyle{\int_{0}^{1}{(1+\frac{x}{2} +\frac{x^2}{3} +\frac{x^3}{4} +\frac{x^4}{5}+... )} }$ $= -1[x+\frac{x^2}{4} +\frac{x^3}{9} +\frac{x^4}{16}+...]_{0}^{1} = -(\displaystyle\sum_{i=1}^{\infty} \frac{1}{i^2}) = \large -\frac{\pi^2}{6}$ The last step comes from the riemann zeta function.

we know that: $\displaystyle \int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$ So using that we can say: $I= \displaystyle \int_{0}^{1} \frac{ln(1-x)}{x} dx= \int_{0}^{1} \frac{ln(x)}{1-x}dx$ Since : $0<x<1$ the infinite sum of a geometric series of the form $u_{n}=x^{n}$ can be written as: $\frac{1}{1-x}$ or:

$\displaystyle \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ Substituting this back to the integral we get: $I = \displaystyle \int_{0}^{1} \sum_{n=0}^{\infty}ln(x)*x^{n} dx$

And using the linearity of the integral we can interchange the integral sign and the sum to get: $I = \displaystyle \sum_{n=0}^{\infty} \int_{0}^{1} x^n*ln(x) dx$

Now all we do is calculate integral by parts and sum it up. So using parts we get: $I = \displaystyle \int_{0}^{1} x^n*ln(x) dx = \frac{-1}{(n+1)^{2}}$

So we get: $I = \displaystyle -\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}$ $I = \frac{-(\pi)^2}{6}$