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

Dec 30, 2016

As the denominator $\left(1 + {x}^{2}\right)$ 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.

Dec 30, 2016

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

$\frac{x - 1}{1 + {x}^{2}} = \frac{1 + i}{2 \left(x - i\right)} + \frac{1 - i}{2 \left(x + i\right)}$

#### 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:

$1 + {x}^{2} = \left(x - i\right) \left(x + i\right)$

So we have:

$\frac{x - 1}{1 + {x}^{2}} = \frac{A}{x - i} + \frac{B}{x + i}$

$\textcolor{w h i t e}{\frac{x - 1}{1 + {x}^{2}}} = \frac{A \left(x + i\right) + B \left(x - i\right)}{1 + {x}^{2}}$

$\textcolor{w h i t e}{\frac{x - 1}{1 + {x}^{2}}} = \frac{\left(A + B\right) x + i \left(A - B\right)}{1 + {x}^{2}}$

Equating coefficients, we have:

$\left\{\begin{matrix}A + B = 1 \\ i \left(A - B\right) = - 1\end{matrix}\right.$

Hence:

$A - B = \frac{- 1}{i} = i$

Adding the first equation to this one, we have:

$2 A = 1 + i$

Hence:

$A = \frac{1}{2} + \frac{1}{2} i$

and:

$B = \frac{1}{2} - \frac{1}{2} i$

So:

$\frac{x - 1}{1 + {x}^{2}} = \frac{1 + i}{2 \left(x - i\right)} + \frac{1 - i}{2 \left(x + i\right)}$