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

May 8, 2016

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

#### Explanation:

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

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

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

As $\left(1 - {x}^{2}\right) = \left(1 - x\right) \left(1 + x\right)$, let

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

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

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

Hence $A + B = 0$ and $A - B = - 2$. Adding them we get

$2 A = - 2$ or $A = - 1$ and $B = 1$

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