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Stuck on an integral

  1. Mar 26, 2007 #1
    1. The problem
    Find the definite integral.
    [tex]\int \frac {t^2 - 3}{-t^3 + 9t + 1} dt[/tex]

    2. The attempt at a solution
    The answer is the book made it seem like the rule [itex] \int \frac {du}{u} = ln|u| + C [/itex] was used. Here's what I got:
    [tex]u = -t^3 + 9t + 1[/tex]
    [tex]du = -3t^2 + 9\;dt[/tex]
    [tex]-\frac{1}{3}du = t^2 + 9\;dt[/tex]
    Obviously this won't work out, so is there something I'm overlooking that anyone can point out?
    Last edited: Mar 26, 2007
  2. jcsd
  3. Mar 26, 2007 #2
    yes that rule is used, but when you take du, you're entire quantity needs to be multiplied by dt.
  4. Mar 26, 2007 #3
    the last line of your calculation is wrong.

    notice that it should be:

    you forgot changing the dt.
  5. Mar 26, 2007 #4
    you've got the right substitution. By rearranging your du formula for dt you'll get:

    dt = -du/(3(t^2-3))
    and it will cancel out leaving you with the simple integral
    Last edited: Mar 26, 2007
  6. Mar 27, 2007 #5
    Wow, I feel dumb. Thanks for pointing that out, though.
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