see the 71st sum of arihant integral calculus if you have it (Page 109, solutions page 391) :)

 

The basic idea is to write [x^4 + 1]/[x^6 + 1] as[(x^2 + 1)^2 - 2x^2)]/[(x^2 + 1)(x^4 - x^2 + 1)]