Question

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

Answer 1

The answer is `=1/(x)+(1/4)/(x-1)^2+(-1/2)/(x-1)+(-1/4)/(x+1)^2+(-1/2)/(x+1)`

Explanation:

Let's rewrite the expression

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

The decomposition into partial fractions is

`1/(x(x-1)^2(x+1)^2)=A/(x)+B/(x-1)^2+C/(x-1)+D/(x+1)^2+E/(x+1)`

`=(A(x-1)^2(x+1)^2+Bx(x+1)^2+Cx(x-1)(x+1)^2+Dx(x-1)^2+Ex(x-1)^2(x+1))/(x(x-1)^2(x+1)^2)`

Therefore,

`1=A(x-1)^2(x+1)^2+B(x)(x+1)^2+Cx(x-1)(x+1)^2+Dx(x-1)^2+Ex(x-1)^2(x+1)`

Let `x=0`, `=>`, 1=A#

Let `x=1`, `=>`, `1=4B`, `=>`, `B=1/4`

Let `x=-1`, `=>`, `1=-4D`, `=>`, `D=-1/4`

Coefficients of `x^4`, `=>`,`0=A+C+E`, `=>`, `C+E=-1`

Coeficients of `x^3`, `=>`, `0=B+C+D-E`, `=>`, `C-E=0`

`C=E=-1/2`

`1/(x(x-1)^2(x+1)^2)=1/(x)+(1/4)/(x-1)^2+(-1/2)/(x-1)+(-1/4)/(x+1)^2+(-1/2)/(x+1)`