Question

How do you express `1/(x^2+x+1)` in partial fractions?

Answer 1

`1/(x^2+x+1)=(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))`

Explanation:

This expression can only be split down further if we use Complex coefficients...

`x^2+x+1`

`= (x+1/2)^2+3/4`

`= (x+1/2)^2-(sqrt(3)/2i)^2`

`= (x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)`

So:

`1/(x^2+x+1) = A/(x+1/2-sqrt(3)/2i)+B/(x+1/2+sqrt(3)/2i)`

`=(A(x+1/2+sqrt(3)/2i)+B(x+1/2-sqrt(3)/2i))/(x^2+x+1)`

`=((A+B)x+(A(1/2+sqrt(3)/2i)+B(1/2-sqrt(3)/2i)))/(x^2+x+1)`

Equating coefficients, we find:

`{ (A+B=0), (A(1/2+sqrt(3)/2i)+B(1/2-sqrt(3)/2i)=1) :}`

From the first equation we find `B=-A`.

Substitute this in the second equation to get:

`Asqrt(3)i = 1`

Hence:

`{ (A = 1/(sqrt(3)i) = -sqrt(3)/3i), (B=sqrt(3)/3i) :}`

So:

`1/(x^2+x+1) = B/(x+1/2+sqrt(3)/2i)+A/(x+1/2-sqrt(3)/2i)`

`=(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))`