Sum this series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\ldots$ upto $n$ terms 
Sum this series: $$\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\ldots$$ upto $n$ terms.  

My approach:
$$(1-n^6)=(1-n^2)(1+n^2+n^4)\implies \dfrac{n}{1+n^2+n^4}=\dfrac{n(1-n^2)}{1-n^6}$$  
So, the above series can be written as $$\sum\limits_{i=1}^n \dfrac{i(1-i^2)}{1-i^6}$$  
I suppose that this can now be converted into integration which I cannot apparently. Please help. It would be better if the solution is not based upon integration but algebra.

HINT:
As $\displaystyle1+n^2+n^4=(1+n^2)^2-n^2=(1+n+n^2)(1-n+n^2)$
$$\frac{2n}{1+n^2+n^4}=\frac{(1+n+n^2)-(1-n+n^2)}{(1+n+n^2)(1-n+n^2)}=\cdots$$
Again if $\displaystyle f(n)=\frac1{n^2-n+1}, f(n+1)=?$
So, we are dealing with a Telescoping series
