Calculate the sum of the series $\sum_{n=1}^\infty \frac 1{n(n+1)^2}$ I have already proven that
$$\int_0^1 \ln(1-x)\ln (x)\mathrm d x=\sum_{n=1}^\infty \frac 1{n(n+1)^2}.$$
So now I want so calculate

$$S:=\sum_{n=1}^\infty \frac 1{n(n+1)^2}.$$

I think I should use the fact that 
$$\sum_{n=1}^\infty \frac 1{n^2}=\frac{\pi^2}6$$
and I know using Wolfram Alpha that $S=2-\frac{\pi^2}6$.
So finally I will get
$$\int_0^1 \ln(1-x)\ln (x)\mathrm d x=2-\frac{\pi^2}6.$$
But how to calculate $S$ ?
 A: Maybe it is interesting to see another way to solve the integral. Recalling the definition of Beta function $$B\left(a,b\right)=\int_{0}^{1}x^{a-1}\left(1-x\right)^{b-1}dx,\, a,b>0
 $$ we have that $$\frac{\partial^{2}}{\partial a\partial b}B\left(a,b\right)=\int_{0}^{1}x^{a-1}\log\left(x\right)\left(1-x\right)^{b-1}\log\left(1-x\right)dx
 $$ but since $$B\left(a,b\right)=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}\tag{1}
 $$ we have that $$\frac{\partial^{2}}{\partial a\partial b}B\left(a,b\right)=B\left(a,b\right)\left(\left(\psi^{\left(0\right)}\left(a\right)-\psi^{\left(0\right)}\left(a+b\right)\right)\left(\psi^{\left(0\right)}\left(b\right)-\psi^{\left(0\right)}\left(a+b\right)\right)-\psi^{\left(1\right)}\left(b\right)\right)
 $$ where $\psi^{\left(0\right)}\left(x\right),\psi^{\left(1\right)}\left(x\right)
 $ are respectively the digamma and trigamma function. So $$\int_{0}^{1}\log\left(x\right)\log\left(1-x\right)dx=\left.\frac{\partial^{2}}{\partial a\partial b}B\left(a,b\right)\right|_{a=1,b=1}
 $$ $$=B\left(1,1\right)\left(\left(\psi^{\left(0\right)}\left(1\right)-\psi^{\left(0\right)}\left(2\right)\right)^{2}-\psi^{\left(1\right)}\left(2\right)\right)
 $$ and now using $(1)$, and $$\psi^{\left(0\right)}\left(n\right)=\sum_{k=1}^{n-1}\frac{1}{k}-\gamma
 $$ $$\psi^{\left(1\right)}\left(n\right)=\sum_{k\geq0}\frac{1}{\left(k+n\right)^{2}}
 $$ we can conclude $$\int_{0}^{1}\log\left(x\right)\log\left(1-x\right)dx=\color{red}{2-\frac{\pi^{2}}{6}}$$ as wanted.
A: we have that 
$$\sum_{n=1}^{m }\frac{1}{n}- \frac{1}{n+1}=\frac{m}{m+1}=1-\frac{1}{m+1}$$ the sum is $1$when the $m\rightarrow \infty$
$$\sum_{n=1}^{\infty }\frac{1}{n(n+1)^2} = \sum_{n=1}^{\infty }(\frac{1}{n}- \frac{1}{n+1}) -\sum_{n=1}^{\infty } \frac{1}{(n+1)^2}=1 -\sum_{n=2}^{\infty } \frac{1}{n^2}$$
$$=1-(\frac{\pi^2}{6}-1)=2-\frac{\pi^2}{6}$$
