How do you integrate #int 1/[(x^3)-1]# using partial fractions?

1 Answer
Oct 31, 2016

The integral is #1/3ln(x-1)-1/6ln(x^2+x+1)-1/(sqrt3)arctan((2x+1)/sqrt3)#

Explanation:

Let's factorise the denominator
#x^3-1=(x-1)(x^2+x+1)#
and the decomposition in partial fractions
#1/(x^3-1)=A/(x-1)+(Bx+C)/(x^2+x+1)#
#1=A(x^2+x+1)+ (Bx+C)(x-1)#
If #x=0##=>##1=A-C#
coefficients of x^2, #0=A+B#
Coefficients of x, #0=A-B+C#
Solving, we found #A=1/3, B=-1/3, C=-2/3#
so #intdx/(x^3-1)=1/3intdx/(x-1)-1/3int((x+2)dx)/(x^2+x+1)#
#intdx/(x-1)=ln(x-1)#
#int((x+2)dx)/(x^2+x+1)=1/2int((2x+1)dx)/(x^2+x+1)+3/2intdx/(x^2+x+1)#
#1/2int((2x+1)dx)/(x^2+x+1)=1/2ln(x^2+x+1)#
#intdx/(x^2+x+1)=intdx/((x+1/2)^2+3/4)#
let's do the substitution,#u=(2x+1)/sqrt3##=>##(du)/dx=2/sqrt3#
#intdx/((x+1/2)^2+3/4)=2/sqrt3int(du)/(u^2+1)=2/sqrt3arctanu#