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.

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 $$\frac{\pi^2}{6}-1$$
