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Homework Help: Integration using Partial Fractions

  1. Feb 27, 2006 #1

    Hootenanny

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    I need to find the following intergral:
    [tex] \int_{0}^{1} \frac{28x^2}{(2x+1)(3-x)} \;\; dx [/tex]
    So I split it into partial fractions thus:
    [tex] \frac{2}{2x+1} + \frac{36}{3-x} - 14 [/tex]
    Then integrated:
    [tex] \int_{0}^{1} \frac{2}{2x+1} + \frac{36}{3-x} - 14 \;\; dx [/tex]
    [tex] = \left[ \ln\left| 2x+1 \right| + 12\ln\left| 2-x \right| - 14x \right]_{0}^{1} [/tex]
    But this isn't going to give me the correct answer which is quoted as:
    [tex]37\ln 3 - 36\ln 2 - 14 [/tex]
    Can anybody see where I've gone wrong? Thank's
     
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  3. Feb 27, 2006 #2

    Hootenanny

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    Sorry just a correction in my answer:
    [tex] = \left[ \ln\left| 2x+1 \right| + 12\ln\left| 3-x \right| - 14x \right]_{0}^{1} [/tex]
     
  4. Feb 27, 2006 #3

    Physics Monkey

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    Your partial fraction expansion is ok. However, you integrated [tex] \frac{36}{3 - x} [/tex] incorrectly.
     
  5. Feb 27, 2006 #4

    Hootenanny

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    Ahh yes. Thank-you, I realise what I've done now.
     
  6. Feb 27, 2006 #5

    Hootenanny

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    Is there any standard intergral in the form of [itex] \int \frac{a}{b+x} = ... [/itex] because at the moment I'm not sure what to do. If anybody could guide me through it?
     
  7. Feb 27, 2006 #6
    I suppose you can use [tex]\int\frac{a}{b\pm x}dx=\pm a\ln|x\pm b|[/tex]

    - Kamataat
     
  8. Feb 27, 2006 #7

    Hootenanny

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    Thank you very much.
     
  9. Dec 29, 2007 #8
    hi there, I was wondering if you could possibly help me with an integration problem


    [tex] \int_{1/4}^{-1/4} \frac{1}{(1-4x^2)} \;\; dx [/tex]

    i managed to split it into partial fractions and hence split up the integration as follows

    [tex] \frac{1}{2} \int_{1/4}^{-1/4} \frac{1}{2x+1} - \frac{1}{2x-1} \;\;dx[/tex]

    after this i integrated it in the following manner but im unsure as to whether it is right, and if so, how to continue it.

    [tex][\frac{1}{2} (\frac{1}{2}\ln(2x+1)-\frac{1}{2}\ln(2x-1))] [/tex]still between the same two limits

    Can anyone offer a hand?
     
  10. Dec 29, 2007 #9
    if my integration was correct please continue reading this post, if not then it doesn't matter!!

    I broke down that inegraton to read

    [tex][\frac{1}{2} \ln\frac{2x+1}{2x-1}] [/tex] limits [tex]\pm\frac{1}{4}[/tex]

    i then simplified this down to the following (including limit values)

    [tex] \frac{1}{2} \ln\frac{1.5}{-0.5} - \frac{1}{2}\ln\frac{0.5}{-1.5} [/tex]

    Using the rules of logs i calculated this to be

    [tex] \ln\frac{-3}{\frac{-1}{3}} [/tex]

    Giving a final answer of [tex]\ln9[/tex]

    Does anyone know if this is the correct answer or have i gone about it in the wrong way?
     
  11. Dec 29, 2007 #10
    The correct answer is

    [tex]\frac{1}{4} \left(\ln|2x+1|-\ln|-2x+1|\right) [/tex]

    you need the absolute values in [tex]\ln(x)[/tex] in order to be well defined.

    See Kamataat's post [tex]\int\frac{a}{b\pm x}dx=\pm a\ln|x\pm b|[/tex]
     
  12. Dec 29, 2007 #11
    well you have [tex]\frac{1}{1-4x^2}[/tex]

    so it should become

    [tex]\frac{1}{(1+2x)(1-2x)}[/tex]

    if you want to write it as (2x+1)(2x-1), then you have to factor out a negative
     
  13. Dec 29, 2007 #12
    you have it the other way around, i even checked it on my calculator and my answer matches:

    after integrating it should become

    [tex]\frac{1}{4}\left[\ln\left|\frac{1-2x}{1+2x}\right|\right]_{.25}^{-.25}[/tex]
     
  14. Dec 29, 2007 #13
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