Have you learned how to solve complex integrals? 'Complexifying' the integral and then using the residue theorem can solve this quite easily, but it won't help if you haven't studied those tools.
This is a truly ugly problem, but it can be solved . . .
The denominator is the sum of cubes: .
. . which factors: .
The quartic can also be factored . . .
Add and subtract
. . . (difference of squares)
The denominator is: .
You know what to do, right?
Apply partial fraction decomposition . . .
. . . . . . . . .
I'll wait in the car . . .
Consider the integral next to :
Sub. ( maybe i have been using this substitution much more times than trigo. sub. since I realized its power and magic ! )
Summation gives :
so we find that the answer is
If i am right , your generalized integral
I took Defunkt's advised (See the attached pdf): Complexifying integral.pdf