Let, I=int(xdx)/(x^3-1)=intx/{(x-1)(x^2+x+1)}dx.I=∫xdxx3−1=∫x(x−1)(x2+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.!
