Question

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

Answer 1

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`