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Homework Help: Integration Help - Which Method?

  1. Feb 4, 2014 #1

    i'm having some trouble identifying the correct method to approach a problem. I don't have any working to show, so i'll explain why I think the methods i've considered are not correct.

    1. The problem statement, all variables and given/known data

    Integrate with respect to [itex]x[/itex] the following functions.

    [itex]\int \frac{x}{(3-x)^{7}} dx[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I think the method of changing the variable is not suitable here because I can't see a link between the derivative of the denominator and the numerator.

    Similarly I don't think it is possible to use the method that identifies the numerator as the derivative of the denominator allowing you to say the integral is equal to ln|f(x)| + c

    I'd really appreciate a nudge in the right direction!

    Thank you,

  2. jcsd
  3. Feb 4, 2014 #2


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    Use the substitution u=3-x.

  4. Feb 4, 2014 #3
    I'm struggling to make this work.

    [itex]\int \frac{x}{(3-x)^{7}} dx[/itex]

    Let [itex] u \equiv (3 - x)[/itex]

    [itex]\frac{du}{dx} = -1[/itex]

    [itex] ... du \equiv ... -1dx[/itex]

    [itex]\int \frac{x}{(3-x)^{7}} dx = -x \int \frac{du}{u^{7}}[/itex]

    (that seems wrong to me, I don't think we've ever been told it's ok to bring an x outside the integral sign)

    I get;

    [itex]\int \frac{x}{(3-x)^{7}} dx = \frac{x}{6(3-x)^6} + c[/itex]
  5. Feb 4, 2014 #4


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    You can't just pull out the x like you did. You have to replace the x in the numerator with x=3-u and leave it under the integral sign.
  6. Feb 4, 2014 #5
    The item in red is wrong . You need to express 'x' present in the Nr also in terms of 'u' .After the substitution ,you should have ##\int \frac{u-3}{u^7} du##
  7. Feb 4, 2014 #6

    Ray Vickson

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    Others have suggested a change of variables, but if you do straight integration by parts you can bypass that: just set ##u = x, \: dv = dx/(3-x)^7 = (3-x)^{-7} \, dx##.
  8. Feb 4, 2014 #7


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    Think again about your substitution: if [itex] u = 3 -x [/itex] then [itex] x = 3 - u [/itex]. How will that help your integrand
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