Question

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

Answer 1

`int1/(x^3-1) dx=1/6[2ln|x-1| -ln|x^2+x+1|-2sqrt3arctan(((2x+1)sqrt3)/3)]+"c"`

Explanation:

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"`