Suppose $$\frac{2n+1}{n^2(n+1)^2} = \frac{A}{n} + \frac{B}{n+1}+\frac{C}{n^2} + \frac{D}{(n+1)^2}$$ Then we have $$2n+1=An(n+1)^2 + Bn^2(n+1) +C(n+1)^2 +Dn^2$$ Let $n=0 \implies C = 1$ and Let $n=-1 \implies D = -1$. Now we note that $$C(n+1)^2 - Dn^2 = (n+1)^2 - n^2 = 2n+1$$ $$\therefore\quad A = B = 0\implies\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2} $$ Hence, $$\sum_{n=1}^{\infty}\frac{2n+1}{n^2(n+1)^2} = \Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)+\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)+\cdots = \boxed{1}$$