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Integral question

  1. Jun 29, 2005 #1
    Here is the problem:

    [tex] \int ^2_1 \frac{dx}{(3-5x)^2} [/tex]

    [tex] u = 3-5x [/tex]

    [tex] du = -5dx [/tex]

    [tex] dx = -\frac{1}{5} du [/tex]

    so, (with the 7 and 2 being negatives)

    = [tex] -\frac{1}{5} \int^7_2 \frac{du}{u^2} [/tex]

    = [tex] -\frac{1}{5} (-\frac{1}{u})]^7_2 [/tex]

    But what I don't understand is how the [tex] u^2 [/tex] becomes just [tex] u [/tex].
  2. jcsd
  3. Jun 29, 2005 #2
    Why did you change your bounds? You had everything right.

    [tex]\frac{1}{u^2}[/tex] can be written as [tex]u^{-2}[/tex]

    Now follow the general power rule and [tex]\int u^{-2}du = \frac{(u^{-1})}{(-1)}[/tex]

    But to write that more nicely write it as [tex]-\frac{1}{u}[/tex]

    Make sense?
    Last edited: Jun 29, 2005
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