Let, #I=int(xdx)/(x^3-1)=intx/{(x-1)(x^2+x+1)}dx.#
Let us decompoose the Integrand using the Partial Fraction as
#x/{(x-1)(x^2+x+1)}=A/(x-1)+(Bx+C)/(x^2+x+1); A,B,C in RR.#
We use Heavyside's Method to find, #A,B,C.#
For #A,# we cover #(x-1)# from the integrand and plug in #x=1# in its
remaining portion, denoted as, #[x/(x^2+x+1)]_(x=1).#
#:. A=[x/(x^2+x+1)]_(x=1)=1/3.#
#:. x/{(x-1)(x^2+x+1)}-(1/3)/(x-1)=(Bx+C)/(x^2+x+1), or,#
#{x-1/3(x^2+x+1)}/{(x-1)(x^2+x+1)}=(Bx+C)/(x^2+x+1).#
#:. (-1/3x^2+2/3x-1/3)/{(x-1)(x^2+x+1)}=(Bx+C)/(x^2+x+1).#
#:. {(cancel(x-1))(-1/3x+1/3)}/{(cancel(x-1))(x^2+x+1)}=(Bx+C)/(x^2+x+1).#
#:. B=-1/3, C=1/3.#
Therefore, #I=1/3int1/(x-1)dx+int(-1/3x+1/3)/(x^2+x+1)dx#
#=1/3ln|x-1|-1/3int(x-1)/(x^2+x+1)dx#
#=1/3ln|x-1|-(1/3)(1/2)int(2x-2)/(x^2+x+1)dx#
#=...,,...-1/6int{(2x+1)-3}/(x^2+x+1)dx#
#=...,,...-1/6int(2x+1)/(x^2+x+1)dx+3/6int1/{(x+1/2)^2+(sqrt3/2)^2}dx#
#=...,,...-1/6ln(x^2+x+1)+(1/2)(1/(sqrt3/2))arc tan{(x+1/2)/(sqrt3/2)}.#
Note that the later integrals were found using the Results :
# (1) : int(f'(x))/f(x)dx=ln|f(x)+c.#
#(2) : int1/{(x+a)^2+b^2}dx=1/b arc tan((x+a)/b)+c' (bne0).#
Finally, we have,
#I=1/3ln|x-1|-1/6ln(x^2+x+1)+1/sqrt3 arc tan ((2x+1)/sqrt3)+C.#
Enjoy Maths.!
