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Integral and Partial fractions

  • Thread starter jimen113
  • Start date
  • #1
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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)
so...
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
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
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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:

[tex]\frac{A}{x-2}+\frac{B}{x-1}+\frac{C}{(x-1)^2}+\frac{D}{x+1}+\frac{E}{(x+1)^2}=\frac{1}{(x-2)(x-1)^2(x+1)^2}[/tex]

right?

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

[tex]A(x-1)^2(x+1)^2+B(x-2)(x-1)(x+1)^2+C(x-2)(x+1)^2+D(x-2)(x-1)^2(x+1)+E(x-2)(x-1)^2=1[/tex]

Now set [itex]x=1 \Rightarrow C=-\frac{1}{4}[/itex], [itex]x=2 \Rightarrow A=\frac{1}{9}[/itex] etc.
 
Last edited:
  • #3
zcd
200
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You made the mistake over here:
Code:
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
67
0
Thank you so much for your help zcd and cyosis. Last question: I get A= (1/9) C=(-1/4)
E=(-1/12)
But to find D and B I tried plugging in zero and didn't get anywhere. What other numbers can I plug in to solve for D and E? :blushing:
 
  • #5
Cyosis
Homework Helper
1,495
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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.
 
Last edited:
  • #6
67
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omg...thank you for your help...
I did try plugging in (-2) and I end up with:
1+B(12)+1+D(36)+3=1
and B(12)+D(36)=-4
so I'm still left with the two unknowns...
 
  • #7
Cyosis
Homework Helper
1,495
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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
67
0
thanks.
 

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