Proving that $\frac{3}{2} \sum_{k=1}^{\infty} \frac{4}{k^3+k^2} = \pi^2-6$ I'm trying to prove that:
$$\frac{3}{2} \sum_{k=1}^{\infty} \frac{4}{k^3+k^2} = \pi^2-6$$
I've tried looking at the partial sums, but no luck there. I just have no idea where to begin.
Knowing that $\frac{4}{k^3+k^2} = \frac{4}{k^2 (k+1)} \implies \frac{4}{k^2}+\frac{4}{(k+1)}-\frac{4}{k}$ seemed to help, but I get stuck no matter how I look at it.
Any help would be greatly appreciated.

Hint:
$$
    \frac{4}{k^3+k^2} = \frac{4}{k^2(k+1)} = \frac{a_1}{k^2} + \frac{a_2}{\vphantom{k^2}k} + \frac{a_3}{\vphantom{k^2}k+1}
$$
for some constants $a_1$, $a_2$, and $a_3$, and
$$
    \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}.
$$
