How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x - 1) {(x + 1)}^{2}}$ $1/ [(x-1) ( x+ 1) ^2]$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x - 1) {(x + 1)}^{2}}$ $1/ [(x-1) ( x+ 1) ^2]$ ?

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Answer 1 · 2015-12-07T01:33:28

Decompose into 3 separate terms ...

Explanation:

$\frac{A_{1}}{x - 1} + \frac{A_{2}}{x + 1} + \frac{A_{3}}{{(x + 1)}^{2}} = \frac{1}{(x - 1) {(x + 1)}^{2}}$ $A_1/(x-1)+A_2/(x+1)+A_3/(x+1)^2=1/((x-1)(x+1)^2)$

Now, get a common denominator ...

$\frac{A_{1} {(x + 1)}^{2} + A_{2} (x - 1) (x + 1) + A_{3} (x - 1)}{(x - 1) {(x + 1)}^{2}} = \frac{1}{(x - 1) {(x + 1)}^{2}}$ $[A_1(x+1)^2+A_2(x-1)(x+1)+A_3(x-1)]/[(x-1)(x+1)^2]=1/((x-1)(x+1)^2)$

Now, set the numerators equal...

$A_{1} {(x + 1)}^{2} + A_{2} (x - 1) (x + 1) + A_{3} (x - 1) = 1$ $A_1(x+1)^2+A_2(x-1)(x+1)+A_3(x-1)=1$

Next, match up the common terms ...

$x^{2}$ $x^2$ terms: $A_{1} + A_{2} = 0$ $A_1+A_2=0$

$x$ $x$ terms: $2 A_{1} + A_{3} = 0$ $2A_1+A_3=0$

constant terms: $A_{1} - A_{2} - A_{3} = 1$ $A_1-A_2-A_3=1$

Finally, with 3 equations and 3 unknowns, solve for $A_{1}, A_{2}, and A_{3}$ $A_1, A_2, and A_3$

$A_{1} = \frac{1}{4}$ $A_1=1/4$

$A_{2} = - \frac{1}{4}$ $A_2=-1/4$

$A_{3} = - \frac{1}{2}$ $A_3=-1/2$

ANSWER :

$\frac{A_{1}}{4 (x - 1)} - \frac{A_{2}}{4 (x + 1)} + \frac{A_{3}}{2 {(x + 1)}^{2}}$ $A_1/[4(x-1)]-A_2/[4(x+1)]+A_3/[2(x+1)^2]$

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x - 1) {(x + 1)}^{2}}$ $1/ [(x-1) ( x+ 1) ^2]$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x - 1) {(x + 1)}^{2}}$ $1/ [(x-1) ( x+ 1) ^2]$ ?