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Math Help - Integral with partial fractions

  1. #1
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    Integral with partial fractions



    so I got


    A= -3 , B = 5 C = 5

    and i got ∫-3/x-4 + 5x+5/x+9 dx for my equation

    my answer came out to be



    I dont know why it is wrong.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    I didn't check the details but one mistake is that it should be -3\ln(x-4), not -3\ln(x+4)
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  3. #3
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    Quote Originally Posted by emurphy View Post


    so I got



    A= -3 , B = 5 C = 5

    and i got ∫-3/x-4 + 5x+5/x+9 dx for my equation



    my answer came out to be


    I dont know why it is wrong.
    As well as the mistake already mentioned (which may well be just a typo), your calculation of \int \frac{5x + 5}{x^2 + 9} \, dx is very wrong. Please show all the details of your calculation so that it can be reviewed.
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    MHF Contributor Bruno J.'s Avatar
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    Oh yeah that is pretty bad. Use parentheses! You wrote 3/x-4 + 5x+5/x^2 + 9 and integrated it as 3/(x-4) + 5x+5/(x^2 + 9) when in fact you should have written 3/(x-4) + (5x+5)/(x^2 + 9) . These expressions are completely different from one another.
    Last edited by mr fantastic; October 19th 2009 at 09:47 PM. Reason: Fixed latex (you need to use ^2 for x^2)
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  5. #5
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    For \int \frac{5x + 5}{x^2 + 9} \, dx

    I separated the fraction into 2

    ∫5x+5 dx + ∫1/x+9 dx

    ∫5x+5 dx = 5x/2+5x

    ∫1/x+9 dx = ∫1/(1/9)(x+1) dx = 9 ∫ 1/x+1
    = 9tan^-1(x)
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  6. #6
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    Quote Originally Posted by emurphy View Post
    For \int \frac{5x + 5}{x^2 + 9} \, dx

    I separated the fraction into 2

    ∫5x+5 dx + ∫1/x+9 dx

    ∫5x+5 dx = 5x/2+5x

    ∫1/x+9 dx = ∫1/(1/9)(x+1) dx = 9 ∫ 1/x+1
    = 9tan^-1(x)
    NO! You cannot seperate the integrand like that!! Using this logic you would solve \int \frac{x}{x} \, dx as \int x \, dx + \int \frac{1}{x} \, dx which is clearly so wrong that I don't need to say anything more.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    NO! You cannot seperate the integrand like that!! Using this logic you would solve \int \frac{x}{x} \, dx as \int x \, dx + \int \frac{1}{x} \, dx which is clearly so wrong that I don't need to say anything more.
    Adding to this, as a diagnostic of where you're at could you please do the following two integrals:

    I_1 = \int \frac{2x}{x^2 + 9} \, dx

    I_2 = \int \frac{5}{x^2 + 9} \, dx


    Edit: If you cannot do these there is little point in you attempting questions like the one you posted until you have thoroughly reviewed the basic integration techniques this questions assumes and have developed a more solid understanding of integration.
    Last edited by mr fantastic; October 20th 2009 at 12:05 AM.
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