Question

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

Answer 1

As the denominator `(1+x^2)` is a quadratic polynomial, which cannot be factorized and numerator is already a linear binomial, of degree less than that of denominator, it is already in partial fractions.

Answer 2

This is already in irreducible form using Real coefficients, but with Complex coefficients we have:

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

Explanation:

If we stick with Real coefficients, then this rational expression is already in irreducible form: The denominator has no linear factors with Real coefficients and the numerator is linear.

If we use Complex coefficients, then we can factor the denominator as:

`1+x^2 = (x-i)(x+i)`

So we have:

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

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

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

Equating coefficients, we have:

`{ (A+B = 1), (i(A-B) = -1) :}`

Hence:

`A-B = (-1)/i = i`

Adding the first equation to this one, we have:

`2A = 1+i`

Hence:

`A=1/2+1/2i`

and:

`B=1/2-1/2i`

So:

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