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Math Help - Integrate [(x^3+x^2-4x)/(x^2-4x+4)] by parts.

  1. #1
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    Integrate [(x^3+x^2-4x)/(x^2-4x+4)] by parts.

    Hello.

    I know I have done something very wrong, but not sure where.
    We have just learnt how to integrate by parts.

    I have uploaded a scan of my working in this link :
    http://imageshack.us/m/842/448/mhf.png
    (It gave me an error when I tried to insert image into this post)

    Thank you very much .
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  2. #2
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    You don't use integration by parts. Here, since the numerator is of greater degree than the denominator, you first have to long divide.

    \displaystyle \frac{x^3 + x^2 - 4x}{x^2 - 4x + 4} = x + 5 + \frac{12x - 20}{x^2 - 4x + 4}.

    Then you need to factorise the denominator of the proper fraction and apply the partial fractions technique.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by anees7112 View Post
    Hello.

    I know I have done something very wrong, but not sure where.
    We have just learnt how to integrate by parts.

    I have uploaded a scan of my working in this link :
    http://imageshack.us/m/842/448/mhf.png
    (It gave me an error when I tried to insert image into this post)

    Thank you very much .
    With the substitution t=x-2 function to be integrated becomes...

    \frac{x^{3} + x^{2} - 4 x}{x^{2}-4 x +4}= t+7+\frac{12}{t}+ \frac{4}{t^{2}} (1)

    Kind regards

    \chi \sigma
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  4. #4
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    Thank you both for the answer. I am able to do the question now
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