In #(13x+2)/(x^3-1)#, we should first factorize #x^3-1#, using identity #(a^3+b^3)=(a+b)(a^2-ab+b^2)#.
Hence factors are given by #(x^3-1)=(x-1)(x^2-x+1)#.
Partial fractions of #(13x+2)/(x^3-1)#, will be
#(13x+2)/(x^3-1)hArrA/(x-1)+(Bx+C)/(x^2+x+1)#and simplifying RHS it becomes
#(13x+2)/(x^3-1)hArr(A(x^2+x+1)+(Bx+C)(x-1))/((x-1)(x^2+x+1))# or
#(13x+2)/(x^3-1)hArr((Ax^2+Ax+A)+(Bx^2-Bx+Cx-C))/(x^3-1)# or
#(13x+2)/(x^3-1)hArr((A+B)x^2+(A-B+C)x+(A-C))/(x^3-1)#
Hence, we have #A+B=0# - (i), #A-B+C=13# - (ii) and #A-C=2# - (iii).
From (i) we get #B=-A#, from (iii) #C=A-2# and putting them in (ii),
we get #A-(-A)+(A-2)=13# or #3A=15# or #A=5#
Hence #B=-5# and #C=3#
Hence partial fractions are #5/((x-1))+(-5x+3)/(x^2+x+1)#
