How to integrate the dilogarithms? $\def\Li{\operatorname{Li}}$
How can you integrate $\Li_2$? I tried from $0 \to 1$
$\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$
$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$
$$(An + B)(n+1) + D(n^2) = 1$$
Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$
$$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$
$$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$ 
The $1/n$ is the problem, it is the harmonic series, which diverges.
 A: Maybe you should look at your decomposition as
$$\frac1{n^2 (n+1)} = \frac1{n^2} - \frac1{n (n+1)}$$
The sum over the second term is easy, given that the indefinite sum is telescoping, i.e., 
$$\sum_{n=1}^N \frac1{n (n+1)} = 1-\frac1{N+1}$$
We take the limit as $N \to \infty$ and then we may view this as the infinite sum.  (Otherwise, as you say, there are convergence issues.)
Thus the sum in question is $$\frac{\pi^2}{6}-1$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{\rm Li}_{2}\pars{x}\,\dd x}
=\left.\vphantom{\Large A}x\,{\rm Li}_{2}\pars{x}\right\vert_{0}^{1}
-\int_{0}^{1}x\,{\rm Li}_{2}'\pars{x}\,\dd x
\\[5mm]&=\overbrace{{\rm Li}_{2}\pars{1}}
^{\ds{\color{#c00000}{\sum_{n\ =\ 1}^{\infty}{1^{n} \over n^{2}}\ =\
{\pi^{2} \over 6}}}}\ -\ 
\int_{0}^{1}x\ \overbrace{\bracks{-\,{{\ln\pars{1 - x} \over x}}}}
^{\ds{=\ \color{#c00000}{{\rm Li}_{2}'\pars{x}}}}\ \,\dd x
={\pi^{2} \over 6} +\
\underbrace{\int_{0}^{1}\ln\pars{x}\,\dd x}_{\ds{=\ \color{#c00000}{-1}}}
\\[5mm]&=\color{#66f}{\large{\pi^{2} \over 6} - 1} \approx {\tt 0.6449} 
\end{align}
A: $$
\begin{align}
\int_0^1\mathrm{Li}_2(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=1}^\infty\frac{x^k}{k^2}\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\frac1{(k+1)k^2}\\
&=\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1k+\frac1{k+1}\right)\\
&=\zeta(2)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\
&=\frac{\pi^2}{6}-1
\end{align}
$$
where
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\
&=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\[12pt]
&=1
\end{align}
$$
is a telescoping series.

Note that we only break $\left(\dfrac1k-\dfrac1{k+1}\right)$ into $\dfrac1k$ and $\dfrac1{k+1}$ when we are looking at finite sums. When considering infinite sums, it is kept together, and is equal to $\dfrac1{k(k+1)}$.
