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

May 8, 2016

$\frac{{x}^{2} + x + 1}{1 - {x}^{2}} = - 1 + \frac{3}{2 \left(1 - x\right)} + \frac{1}{2 \left(1 + x\right)}$

#### Explanation:

$\frac{{x}^{2} + x + 1}{1 - {x}^{2}} = - 1 + \frac{1 - {x}^{2} + {x}^{2} + x + 1}{1 - {x}^{2}}$ or

$\frac{{x}^{2} + x + 1}{1 - {x}^{2}} = - 1 + \frac{x + 2}{1 - {x}^{2}}$ and let

$\frac{x + 2}{1 - {x}^{2}} = \frac{x + 2}{\left(1 - x\right) \left(1 + x\right)} \Leftrightarrow \frac{A}{1 - x} + \frac{B}{1 + x}$ or

$\frac{x + 2}{\left(1 - x\right) \left(1 + x\right)} \Leftrightarrow \frac{A \left(1 + x\right) + B \left(1 - x\right)}{\left(1 - x\right) \left(1 + x\right)}$ or

$\frac{x + 2}{\left(1 - x\right) \left(1 + x\right)} \Leftrightarrow \frac{\left(A - B\right) x + \left(A + B\right)}{\left(1 - x\right) \left(1 + x\right)}$ or

Hence $A - B = 1$ and $A + B = 2$ or $A = \frac{3}{2}$ and $B = \frac{1}{2}$

Hence $\frac{{x}^{2} + x + 1}{1 - {x}^{2}} = - 1 + \frac{3}{2 \left(1 - x\right)} + \frac{1}{2 \left(1 + x\right)}$