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Integrate this indefinite integral: 1/(x-6)^2 dx

  1. Jan 25, 2006 #1
    I need to integrate this indefinite integral: 1/(x-6)^2 dx

    Here is my work...

    Let u= x-6

    du/dx=1

    so: integral 1/u^2 du

    = 3/u^3 + c (constant)
    =3/(x-6)^3 + c

    Have I gone wrong? And if so where? Thanks
     
  2. jcsd
  3. Jan 25, 2006 #2

    StatusX

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    Be careful. What is the derivative of 1/xn? So what is its integral?
     
  4. Jan 25, 2006 #3
    not sure what the derivative is??? what is it?

    The integral is n lnx so is this correct then

    Here is my work...

    Let u= x-6

    du/dx=1

    so: integral 1/u^2 du

    = 2 lnu + c (constant)
    = 2 ln(x-6) + c

    Have I gone wrong? And if so where? Thanks
     
  5. Jan 25, 2006 #4

    StatusX

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    The derivative of 1/xn is -n/xn+1.
     
  6. Jan 25, 2006 #5

    VietDao29

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    You are almost there. Your last step is wrong...
    Look again at your integral table, there should be something like this:
    [tex]\int x ^ \alpha \ dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \ \alpha \neq -1[/tex]
    [tex]\int \frac{1}{x} \ dx = \int x ^ {-1} \ dx = \ln (x) + C \quad \mbox{this is the case for } \alpha = -1[/tex].
    And we also have:
    [tex]\frac{1}{a ^ m} = a ^ {-m}[/tex]. Do you know this?
    So you have:
    [tex]\int \frac{du}{u ^ 2} = \int u ^ {-2} \ du = ?[/tex].
    You can go from here, right? :smile:
     
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