Question

How do you express `(x^2 + 2x) / (x^3 - x^2 + x - 1)` in partial fractions?

Answer 1

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

Explanation:

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

`color(white)((x^2+2x)/(x^3-x^2+x-1)) = A/(x-1)+(Bx+C)/(x^2+1)`

`color(white)((x^2+2x)/(x^3-x^2+x-1)) = (A(x^2+1)+(Bx+C)(x-1))/(x^3-x^2+x-1)`

`color(white)((x^2+2x)/(x^3-x^2+x-1)) = ((A+B)x^2+(-B+C)x+(A-C))/(x^3-x^2+x-1)`

Equating coefficients, we find:

`{ (A+B=1), (-B+C=2), (A-C=0) :}`

Adding all three of these equations, the `B` and `C` terms cancel out, leaving us with:

`2A = 3`

Hence `A=3/2`

Then from the first equation, we find:

`B=1-A = 1-3/2 = -1/2`

From the third equation, we find:

`C = A = 3/2`

So:

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