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∫(6/(x^2 - 4x -12).dx=?

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data
    ∫(6/(x^2 - 4x -12).dx


    3. The attempt at a solution

    i went
    6*∫1/(x^2 - 4x -12)
    which i then worked out to be 6*ln(x^2 - 4x -12)/(2x-4) +c
    ie 6*ln u / u'
    where u=x^2 - 4x -12

    im not very good at this. is this correct?
     
  2. jcsd
  3. Mar 31, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No that is not the integral. I imagine you tried making the substitution u= x^2- 4x- 12 so that du= (2x-4). But you cannot just divide by 2x-4 and say that dx= du/(2x-4) because that is a function of x itself- you cannot just move it inside the integral sign as you could a constant. Instead factor x^2- 4x- 12 and use "partial fractions".
     
  4. Mar 31, 2009 #3
    i dont really know what you are saying. can you dumb it down alot more. im a real dummy :)

    my thinking:
    6/(x^2 - 4x -12). 6 is a constant so it can be brought to the other side of the integral.
    now you have 6*∫1/(x^2 - 4x -12)
    i thought that ∫1/(x^2 - 4x -12) is ln (x^2 - 4x -12)/(2x-4)
    i take it this is where im wrong?
    can you explain better?
     
  5. Mar 31, 2009 #4

    Cyosis

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    Homework Helper

    Using your substitution you would get the following integral:

    [tex]
    6 \int \frac{1}{2x-4} \frac{du}{u}
    [/tex]

    You then seem to treat the [tex] \frac{1}{2x-4} [/tex] term as a constant and you put it in front of the integral and integrate with respect to u. The mistake you make here is that x itself depends on u, so x is not a constant with respect to u. So before you integrate you have to rewrite the fraction in terms of u. As a result you can't take the fraction out of the integral.

    As Ivy suggested you want to factor the denominator and then write the integrand as [tex]\frac{A}{x+a}+\frac{B}{x+b}[/tex]. You should be able to determine the constants A,B,a and b and then solve the integral.

    On a different note if you're unsure whether or not you did the integration correctly, it is a good habit to take the derivative of your answer and see if it equals the integrand. If it doesn't you know you have made a mistake.
     
    Last edited: Mar 31, 2009
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