Partial fraction decomposition

  • Thread starter Ry122
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  • #1
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Problem: Find the integral of 2dx/(2x-3) using partial fraction decomposition
Attempt:
A/(2x-3)
Compare coefficients A = 2
[tex] 2 \int (1/(2x-3)) = 2ln(2x-3)[/tex]
This is wrong. Can someone please tell me what I'm doing wrong?
 

Answers and Replies

  • #2
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Use u-substitution. Let u=2x-3.
 
  • #3
rock.freak667
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Problem: Find the integral of 2dx/(2x-3) using partial fraction decomposition
Attempt:
A/(2x-3)
Compare coefficients A = 2
[tex] 2 \int (1/(2x-3)) = 2ln(2x-3)[/tex]
This is wrong. Can someone please tell me what I'm doing wrong?
you don't need partial fractions for this.

just put u=2x-3 and go from there
 
  • #4
565
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This is just a simple problem that I want to apply the technique to and further my understanding of it so that I can begin using it on more difficult questions.
 
  • #5
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Did the problem tell you that you must use partial fraction for this problem? Because in my opinion, it makes no sense to use it.

Also, [tex]\int \frac{1}{2x-3} \neq ln(2x-3)+C[/tex] therefore [tex]2 \int \frac{1}{2x-3} \neq 2ln(2x-3)+C[/tex]
 
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  • #6
rock.freak667
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This is just a simple problem that I want to apply the technique to and further my understanding of it so that I can begin using it on more difficult questions.
well in a sense...when you have a linear factor in the denominator like 2x-3, in the question, it kinda already is a partial fraction.
 
  • #7
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The integral of 1/x is lnx + c
So then shouldnt the integral of
1/2x-3 be ln(2x-3) + c
If not then what is it?
And yes the question did ask me to use partial fractions for this problem.
 
  • #8
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Integrals of the form a+bu

[tex]\int \frac{1}{a+bu}du=\frac{1}{b}ln|a+bu|+C[/tex]

So then shouldnt the integral of
1/2x-3 be ln(2x-3) + c
To check that this is false: if you take the derivative with respect to x of "ln(2x-3) + c", it will not equal [tex]\frac{1}{2x-3}[/tex] since you have to use the chain rule.
 
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  • #9
rock.freak667
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The integral of 1/x is lnx + c
So then shouldnt the integral of
1/2x-3 be ln(2x-3) + c
If not then what is it?
And yes the question did ask me to use partial fractions for this problem.
[tex]\int \frac{1}{2x-3} dx[/tex]


if u=2x-3, then [itex]\frac{du}{dx}=2 \Rightarrow \frac{du}{2}=dx[/itex]

now just put that into the integral
 
  • #10
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Can partial fraction decomposition even be used in this situation to get a correct answer?
 
  • #11
Dick
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The partial fraction decomposition of 1/(2x-3) IS 1/(2x-3). Please believe what people are telling you.
 
  • #12
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How would i do this one:
integral of ((x^2)+1)/(6x-9x^2)
The highest degree of the denominator is equal to the highest degree of the numerator but it has to be greater than it.
 
  • #13
Defennder
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Do polynomial long division first, to make sure the degree of the denominator is greater than that of the denominator.
 
  • #14
565
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Have no idea how to do that. Can you please show me?
 
  • #15
Gib Z
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What, you mean how to divide?

Well, see if you can follow this:

[tex] \frac{x^2 +1}{6x-9x^2} = \frac{-1}{9}\left( \frac{9x^2+9}{9x^2 - 6x} \right)

= \frac{-1}{9}\left( \frac{9x^2 -6x + 6x +9}{9x^2 - 6x} \right)

= \frac{-1}{9}\left( 1 + \frac{6x+9}{9x^2 - 6x} \right)

= \frac{-1}{9}\left( 1+ \frac{2x+3}{3x^2 - 2x} \right)[/tex].

How about now?
 
  • #16
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i understand but why did the sign of only the denominator change in the 1st step and not the numerator?
Edit: nevermind, figured it out.
THanks for your help everyone
 
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  • #17
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http://www.imagehuge.com/images/a7ttelgdzf3crvdsouk.jpg [Broken]
 
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