Show that

$$\sum\_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$

My try:

We split into partial decomposition 

$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$

Setting $n={1\over 2}$, ${-1\over2}$ we have $A={1\over3}$ and $B={-1\over 3}$

Finding C and D is a bit tedious 

I wonder what is the closed form for

$$\sum\_{n=1}^{\infty}{1\over an+b}=F(a,b)?$$

This way is not a good approach. Can anyone help me with a better approach to tackle this problem? Thank you.