Evaluate $s(x)$ for $0Consider the series
$$
s(x)=\sum_{j=1}^{\infty} \frac{x^j+(1-x)^j}{j^2} .
$$
Evaluate $s(x)$ for $0<x<1$. (You may use that $\sum_1^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}$.)
I have proved that $s(x)$ is continues and continuously differentiable in $(0,1)$ as well as uniform convergent by M-Test. But how can I derive the answer which is
$s(x)=\frac{\pi^2}{6}-(\ln x)(\ln (1-x))$
I have also done the following according to the comment
$$
S(x)=\sum_{j=1}^{\infty} \frac{x^j+(1-x)^j.}{j^2}=\sum_{j=1}^{\infty} \frac{\frac{1}{1-x}+\frac{1}{1-(1-x)}}{j^2}
$$
$S(x)$ converge unitormly
so $\int s(x) d x=\sum_{j=1}^{\infty} \frac{1}{j^2} \int \frac{1}{(1-x) x} d x$
$$
=\frac{\pi^2}{6}(\operatorname{In}(x)-\operatorname{In}(1-x))+c
$$
Am I doing something wrong here?
Thanks in advance! :D
 A: Consider the derivative of $\ln(x) \, \ln(1-x)$ which is
$$ \frac{d}{dx} \, \ln(x) \, \ln(1-x) = \frac{\ln(1-x)}{x} - \frac{\ln(x)}{1-x}.$$
Now integrate both sides which gives
$$ \text{Li}_{2}(x) + \text{Li}_{2}(1-x) = c_{0} - \ln(x) \, \ln(1-x),$$
where $\text{Li}_{2}(x)$ is the dilogarithm function.
Setting $x = 0$, or $x=1$ leads to $c_{0} = \text{Li}_{2}(1) = \zeta(2)$ and
$$ \text{Li}_{2}(x) + \text{Li}_{2}(1-x) = \zeta(2) - \ln(x) \, \ln(1-x). $$
Since $$\text{Li}_{2}(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2}$$
then $$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = \zeta(2) - \ln(x) \, \ln(1-x). $$
Process without using any named functions:
By using
\begin{align}
\frac{d}{dx} \, \ln(x) \, \ln(1-x) &= \frac{\ln(1-x)}{x} - \frac{\ln(x)}{1-x} \\
&= \frac{\ln(1-x)}{x} - \frac{\ln(1 - (1-x))}{1-x} \\
&= - \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} + \sum_{n=1}^{\infty} \frac{(1-x)^{n-1}}{n}
\end{align}
and then integrating both sides gives
$$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = c_{0} - \ln(x) \, \ln(1-x). $$
Setting $x = 0$, or $x=1$, it will be shown that $c_{0} = \zeta(2)$ and leads to the series
$$ \sum_{n=1}^{\infty} \frac{x^n + (1-x)^n}{n^2} = \zeta(2) - \ln(x) \, \ln(1-x). $$
