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Integrals, please help

  1. Feb 15, 2006 #1
    how can i find this integral:

    what would be u in this case? ugh im so confused
  2. jcsd
  3. Feb 15, 2006 #2
    [tex] u = 1+x^3 [/tex]
    [tex]\therefore \frac{du}{dx} = 3x^2 [/tex]
    Then integrate as usual.

    by the way, is this pre-calculus?
  4. Feb 23, 2006 #3
    they teach us a little bit about integrals at the end of precalc.
  5. Feb 23, 2006 #4
    thanks guys and this is calc 2. i just get confused with these integrals by parts and substition
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