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

##### 2 Answers

As the denominator

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))#