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Please help me with this integral

  1. Jun 16, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex]\int \dfrac{x-1}{(x+1)^3} e^x dx[/itex]

    3. The attempt at a solution
    The most I can do is split the terms and integrate them individually but I am facing problems integrating the individual terms.
     
  2. jcsd
  3. Jun 16, 2013 #2
    Do you know about the following:
    [tex]\int e^x(f(x)+f'(x))=e^xf(x)+C[/tex]?
    The problem becomes really easy with that.
     
  4. Jun 17, 2013 #3

    utkarshakash

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    I already know this but I'm finding it difficult to reduce the question to this form. Can you please give me some hints.
     
  5. Jun 17, 2013 #4

    utkarshakash

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    I already know this but I'm finding it difficult to reduce the question to this form. Can you please give me some hints?
     
  6. Jun 17, 2013 #5

    Curious3141

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    Pranav's hint is a good one. But to apply it, you need to see that ##x-1 = (x+1) - 2##. Split that rational expression into two, simplify, and see where you go from there.
     
  7. Jun 17, 2013 #6

    HallsofIvy

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    Thinking you can "split the terms and integrate them individually" indicates that you need to review integration all together!
     
  8. Jun 17, 2013 #7

    utkarshakash

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    This means according to you this step is incorrect.

    [itex]\displaystyle \int \dfrac{(x+1)-2}{(x+1)^3} e^x dx \\
    \displaystyle \int (x+1)^{-2} e^x dx - \int \dfrac{2}{(x+1)^3} e^x dx [/itex]
     
  9. Jun 17, 2013 #8

    CAF123

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    That step is correct. I think what HallsofIvy thought you meant by your statement was that you were going to integrate each 'component' of your integrand separately, ie integrate (x-1), 1/(x+1)3 and ex separately to obtain something completely nonsensical.
     
    Last edited: Jun 17, 2013
  10. Jun 17, 2013 #9

    Curious3141

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    Correct, but you don't actually have to split it up this way. Just leave the integrand as e^x times that expression, then use Pranav's hint.
     
  11. Jun 18, 2013 #10

    utkarshakash

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    I already know that. But HallsOfIvy thought splitting the terms like this is completely wrong.
     
  12. Jun 18, 2013 #11

    Ray Vickson

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    No, he did not think or say that! He just said you need to review integration. Sometimes splitting up an integral is not helpful, even though it may be correct.
     
  13. Jun 24, 2013 #12
    Plain and simple. Write x-1=(x+1)-2 and hence break the denominator. Now you get two separate integrands. Think. Integrating any one of the two by parts should cancel other automatically and you might be able to get the correct integral.

    And do not blame others. :rolleyes:
     
  14. Jun 25, 2013 #13

    utkarshakash

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    I am not blaming others. I have already arrived at the answer.
     
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