A question from G.Polya A question from G.Polya:

prove $$1 + \frac 1 4 + \frac 1 9 + \frac 1 {16} + \cdots 
=  \log x \cdot \log (1 - x) + \frac {x+(1-x)} {1} + \frac {x^2 + (1-x)^2}{4}+\frac{x^3+(1-x)^3}{9}+\cdots$$
for $0&lt;x&lt;1$.

The answer says:

$\sum_{n=1}^\infty \frac{x^n}{n^2}=\int_0^x \sum_{n=1}^\infty \frac{t^{n-1}}{n}dt = - \int_0^{x} t^{-1}\log(1-t)dt;$
integrate by parts then introduce as new variable of integration $s=1-t$.

Following the answer, I got $$- \int_0^{x} t^{-1}\log(1-t)dt=-\log(1-x)\log x+\int _1^ {1-x}\frac{\log(1-t)}tdt.$$
But how to proceed?
Start from your last eqution:

$- \int_0^{x} t^{-1}\log(1-t)dt=-\log(1-x)\log x+\int _1^ {1-x}\frac{\log(1-t)}tdt$

Easy to realize that

$-Li_2(x)=-\ln x ln(1-x)+Li_2(1-x)$ 

where $Li_2(x)$ is polylogarithm function (Spence-function).

$Li_2(x)+Li_2(1-x)=\frac{\pi^2}{6}- \ln x ln(1-x)$

is one of the known $Li_2$ identities 

As  $\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$  we are ready.
Start from your last eqution:

$- \int_0^{x} t^{-1}\log(1-t)dt=-\log(1-x)\log x+\int _1^ {1-x}\frac{\log(1-t)}tdt$

Easy to realize that

$-Li_2(x)=-\ln x ln(1-x)+Li_2(1-x)$ 

where $Li_2(x)$ is polylogarithm function (Spence-function).

$Li_2(x)+Li_2(1-x)=\frac{\pi^2}{6}- \ln x ln(1-x)$

is one of the known $Li_2$ identities 

As  $\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$  we are ready.