# How do you write the partial fraction decomposition of the rational expression 1/ [(x-1) ( x+ 1) ^2]?

Dec 7, 2015

Decompose into 3 separate terms ...

#### Explanation:

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

Now, get a common denominator ...

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

Now, set the numerators equal...

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

Next, match up the common terms ...

${x}^{2}$ terms: ${A}_{1} + {A}_{2} = 0$

$x$ terms: $2 {A}_{1} + {A}_{3} = 0$

constant terms: ${A}_{1} - {A}_{2} - {A}_{3} = 1$

Finally, with 3 equations and 3 unknowns, solve for ${A}_{1} , {A}_{2} , \mathmr{and} {A}_{3}$

${A}_{1} = \frac{1}{4}$

${A}_{2} = - \frac{1}{4}$

${A}_{3} = - \frac{1}{2}$

${A}_{1} / \left[4 \left(x - 1\right)\right] - {A}_{2} / \left[4 \left(x + 1\right)\right] + {A}_{3} / \left[2 {\left(x + 1\right)}^{2}\right]$