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.

$$
\begin{align}
\int_0^1\mathrm{Li}_2(x)\,\mathrm{d}x
&=\int_0^1\sum_{k=1}^\infty\frac{x^k}{k^2}\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\frac1{(k+1)k^2}\\
&=\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1k+\frac1{k+1}\right)\\
&=\zeta(2)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\
&=\frac{\pi^2}{6}-1
\end{align}
$$
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 $\dfrac1{k(k+1)}$.
