In this problem, all numbers , for which  x^5 = 1,  are five different complex roots of equation

     = 1,  k = 1, 2, 3, 4, 5.    (1)

including real root x = 1.  



THREFORE, the numbers   =   are the five complex roots of equation 
     = 1.    (2)

including real number w = 2, which corresponds to real numer 1, which is the root of (1).



Hence, the reciprocal numbers   are the roots of this equation
     = 1.    (3)

including real number w = 1/2, which corresponds to real numer 1, which is the root of (1).



Equation (3) can be written in equivalent polynomial form

     = w^5.    (4)

or

    1 - 5w + 10w^2 - 10w^3 + 5w^4 - w^5 = w^5,

or

    2w^5 - 5w^4 + 10w^3 - 10w^2 + 5w - 1 = 0.    (5)



So, the sum of four addends

     +  +  +     (6)

is the sum of all  four complex roots of equation (5), that are not real numbers.



Let's add    to the sum (6).   We will get the sum  S

    S =  +  +  +  +.    (7)


Now this sum (7) is the sum of all complex roots of equation (5).


According to Vieta's theorem,  this sum (7) is the coefficient at  divided by the coefficient at x^5 
of equation (5), taken with the opposite sign.


This ratio of the mentioned coefficients in (5) is   = .


Hence, the sum (7) is equal to   -  = .


But the problem asks about S - ,  and we finally find this value   

    S -  =  +  +  + =  -  =  = 2.


At this point, the problem is solved completely, to the very end.


ANSWER.  If x⁵ = 1 with x ≠ 1,  then the sum  1/(1+x²) + 1/(1+x⁴) + 1/(1+x) + 1/(1+x³)  is  2.