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Integration help

  1. Mar 27, 2007 #1

    Can you help with the following integral:

    [tex] -\int \frac {2(1+x)} {1-2x-x^2} dx[/tex]

    I'm reasonably sure my setup is correct up to this integral. I tried to factor and do some canceling. - no luck'

    thoughts and direction

    Last edited: Mar 27, 2007
  2. jcsd
  3. Mar 27, 2007 #2
    Have you tried substitution?
  4. Mar 27, 2007 #3


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    Put, x^2+2x-1=z
  5. Mar 27, 2007 #4
    AHHH!! thanks -

    [tex] u = (1-2x-x^2) [/tex]
    [tex] du = -2 - 2x dx[/tex]
    [tex] dx = \frac {du} {-2(1+x)} [/tex]
    [tex] -\int \frac {-du} {u} [/tex]

    [tex] = ln(1-2x-x^2) [/tex]

    This solution is in the exponent of "e"

    and leads to the integral below.
    Question: can you suggest a start for:

    [tex] \int \frac {1-2x-x^2} {(x+1)^2} dx [/tex]

    I've tried various substitutions again and don't see it.

    I've tried [tex] u = -x^2 - 2x [/tex]
    [tex] du = -2x - 2 dx [/tex]
    [tex] (-2(x+1) )dx [/tex]
    [tex] dx = \frac {du} {-2(x+1)}[/tex]

    leaves me with a (x+1) term

    Last edited: Mar 27, 2007
  6. Mar 27, 2007 #5
    Note that [tex]1-2x-x^2 = -(x+1)^2 + 2[/tex].

    Now separate, and integrate.

    This is completing the square. Also you could multiply the bottom out and long divide to get a similar result.
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