The proposition
# (x^4+1)/(x^5+6 x^3)=(x^4+1)/(x^3(x^2+6))=c_1/x+c_2/x^2+c_3/x^3+c_4/(x+i sqrt(6))+c_5/(x-i sqrt(6))#
The #c_k# can be determined according to some techniques. The most elementar is
1) Performing the fractions addition
# (x^4+1)/(x^5+6 x^3) =(1 - 6 c_3 -6 c_2x -(6 c_1 + c_3)x^2+ (i sqrt[6] c_4 - i sqrt[6] c_5-c_2)x^3+(1 - c_1 - c_4 - c_5)x^4)/(x^3 (6 + x^2))#
2) Choosing #c_k# such that
#x^4+1= (1 - 6 c_3 -6 c_2x -(6 c_1 + c_3)x^2+ (i sqrt[6] c_4 - i sqrt[6] c_5-c_2)x^3+(1 - c_1 - c_4 - c_5)x^4), forall x in RR#
3) Solving the conditions
#{(1 - 6 c_3=0), (-6 c_2=0), (6 c_1 + c_3=0), (-c_2 + i sqrt[6] c_4 -
i sqrt[6] c_5=0), (1 - c_1 - c_4 - c_5=0):}#
obtaining
#c_1 = -1/36, c_2 = 0, c_3 = 1/6, c_4 = 37/72, c_5 = 37/72#
#(x^4+1)/(x^5+6 x^3)=1/(6 x^3) - 1/(36 x) + 37/(72 ( x-i sqrt[6] )) + 37/(
72 (x+i sqrt[6]))#
Note. We can avoid the complex expansion doing
#(c'_3x+c'_4)/(x^2+6)# instead of #c_4/(x+i sqrt(6))+c_5/(x-i sqrt(6))#
