Re: partial fraction

• To: mathgroup at smc.vnet.net
• Subject: [mg32881] Re: [mg32865] partial fraction
• From: BobHanlon at aol.com
• Date: Mon, 18 Feb 2002 05:22:00 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```In a message dated 2/16/02 4:45:52 AM, kyguan at hotmail.com writes:

>Do you have any ideas on how to integrate the following function by partial
>
>fraction?
>
>1/(1-x^2)^2 (Sorry it was supposed to be a minus instead of plus)
>
>I have found the answer by using 'the integrator'. However, I have no single
>
>clue on how the answer were calculated.
>

Although it does not appear to be documented, the standard add-on
package DiscreteMath`RSolve` defines the function PartialFractions.

Needs["DiscreteMath`RSolve`"];

f[x_] := 1/(1-x^2)^2;

pfe1 = PartialFractions[f[x], x]

1/(4*(-1 + x)^2) - 1/(4*(-1 + x)) + 1/(4*(1 + x)^2) +
1/(4*(1 + x))

Integrating term-by-term

Integrate[#,x]& /@ pfe1

-1/(4*(-1 + x)) - 1/(4*(1 + x)) - Log[-1 + x]/4 +
Log[1 + x]/4

%//Simplify

((-2*x)/(-1 + x^2) - Log[-1 + x] + Log[1 + x])/4

You could also do the partial fraction expansion yourself

Factor[f[x]]

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

The partial fraction is then of the form

pfe2 = {a, b, c, d}.(1/{x+1, (x+1)^2 , (x-1), (x-1)^2})

d/(-1 + x)^2 + c/(-1 + x) + b/(1 + x)^2 + a/(1 + x)

(cl =

CoefficientList[Numerator[Together[pfe2]],x])==
CoefficientList[Numerator[f[x]],x],
Length[cl]]]

{a + b - c + d == 1, -a - 2*b - c + 2*d == 0,

-a + b + c + d == 0, a + c == 0}

soln = Solve[eqn, {a,b,c,d}][[1]]

{a -> 1/4, b -> 1/4, c -> -1/4, d -> 1/4}

(pfe2 /. soln) == pfe1

True

Bob Hanlon
Chantilly, VA  USA

```

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