Integral and Partial fractions

In summary, the person is trying to solve for A, B, and D using the method of partial fractions. They get confused when they try to plug in negative numbers and end up with an extra factor (x+1) in front of everything. They need help solving for the two unknowns.f
  • #1

Homework Statement

use the method of partial fractions on [tex]\int[/tex] [tex]\frac{36}{(x-2)(x-1)^2(x+1)^2}[/tex] dx

2. Homework Equations

3. The Attempt at a Solution

[tex]\frac{A}{(x-2)}[/tex] + [tex]\frac{B}{(x-1)}[/tex] + [tex]\frac{C}{(x-1)^2}[/tex] + [tex]\frac{D}{(x+1)}[/tex] + [tex]\frac{E}{(x+1)^2}[/tex]
I took the 36 out of the integral. Then

A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1
In the first fraction (A) (all terms cancel)
I tried to plug in x=
1, -1, 0, 2, -2, into (A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1)
when x=1 I ended up with B(-2) +C(-4)=1 (I know this is probably simple algebra or I'm definitely screwing up somewhere

Any help is appreciated
  • #2
Why do you have this extra factor (x+1) in front of everything? For example for A you would get:

[tex]\frac{A}{x-2} (x-2)(x-1)^2(x+1)^2=A (x-1)^2(x+1)^2[/tex]. You have added an extra (x+1) and you're doing that for every term it seems.

You start out with:



Then multiply both sides with the denominator of the right hand side.


Now set [itex]x=1 \Rightarrow C=-\frac{1}{4}[/itex], [itex]x=2 \Rightarrow A=\frac{1}{9}[/itex] etc.
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  • #3
You made the mistake over here:
A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1

Because you multiple the separate fractions by the common denominator, it would end up as

A(x-1)^2 (x+1)^2 + B(x-2)(x-1)(x+1)^2 + ... + E(x-2)(x-1)^2 = 1
  • #4
Thank you so much for your help zcd and cyosis. Last question: I get A= (1/9) C=(-1/4)
But to find D and B I tried plugging in zero and didn't get anywhere. What other numbers can I plug into solve for D and E? :blushing:
  • #5
You will have to pick two values such that B and D aren't multiplied by zero. Since you know all the other constants you will get two equations relating B to D. Solve them and you're good to go.
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  • #6
omg...thank you for your help...
I did try plugging in (-2) and I end up with:
and B(12)+D(36)=-4
so I'm still left with the two unknowns...
  • #7
Try x=0 this will give you 1/9+2B+1/2-2D+1/6=1, -> B-D=1/9. You now have two equations and two unknowns so it is solvable.
  • #8

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