We need to find the partial fraction expansion of #1/(x^3-1)#.
Firstly, we need to rewrite #x^3-1# as a product of irreducible polynomials. #x^3-1=(x-1)(x^2+x+1)#. It's easy to factorise this function since we know #(x-1)# is a root. Then we can use inspection or polynomial division to find the other function.
#therefore1/(x^3-1)=1/((x-1)(x^2+x+1))=A/(x-1)+(Bx+C)/(x^2+x+1)#
Since the numerator has to have a polynomial to a power one less than the denominator, the second term will have to have an #x# in the numerator.
#1=A(x^2+x+1)+(Bx+C)(x-1)#
#A+B=0rArrA=-B# #(1)#
#A-B+C=0# #(3)#
#A-C=1rArrC=A-1# #(2)#
Sub #(1)# and #(2)# into #(3)#
#A+A+A-1=0#
#3A=1#
#A=1/3#
#B=-1/3#
#C=1/3-1=-2/3#
#therefore1/((x-1)(x^2+x+1))=1/(3(x-1))+(-1/3x-2/3)/(x^2+x+1)#
#=1/(3(x-1))-(x+2)/(3(x^2+x+1))#
#int1/(3(x-1)) -(x+2)/(3(x^2+x+1))# #dx#
#=1/3ln|x-1| +"c"_1-1/3int(x+2)/(x^2+x+1)# #dx#
#(x+2)/(x^2+x+1)
= 1/2((2(x+2))/(x^2+x+1))#
#= 1/2((2x+1)/(x^2+x+1) + 3/(x^2+x+1))#
#therefore-1/3int(x+2)/(x^2+x+1)# #dx#
#=-1/6int(2x+1)/(x^2+x+1) + 3/(x^2+x+1)# #dx#
#=-1/6ln|x^2+x+1|+"c"_2-1/6int3/(x^2+x+1)# #dx#
#-1/6int3/(x^2+x+1)# #dx=-3/6int1/((x+1/2)^2+3/4)# #dx#
Let #s=x+1/2# and #ds=dx#
#-1/2int1/((x+1/2)^2+3/4)# #dx=-1/2int1/(s^2+3/4)# #ds#
#=-1/2int(4/3)/(4/3(s^2+3/4))# #ds=-2/3int1/(4/3s^2+1)# #ds#
Let #p=2/sqrt3 s# and #dp=2/sqrt3ds#
#-2/3int1/(4/3s^2+1)# #ds=-1/sqrt3int1/(p^2+1)# #dp#
#=-1/sqrt3arctanp+"c"_3=-sqrt3/3arctan(((2x+1)sqrt3)/3)+"c"_3#
Combining all three integrals gives
#int1/(x^3-1) # #dx=1/3ln|x-1| -1/6ln|x^2+x+1|#
#-sqrt3/3arctan(((2x+1)sqrt3)/3)+"c"#
#=1/6[2ln|x-1| -ln|x^2+x+1|-2sqrt3arctan(((2x+1)sqrt3)/3)]+"c"#
