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Determine the integral

  1. May 24, 2008 #1
    Determine the integral (x^2) / (2x^3 - 3)^2

    This is probably easier than I expect but I have been attempting it using substitution...?
    u = (2x^3 -3)
    du/dx = 6x^2
    dx = du / 6x^2

    x^2 / (u^2) . du / 6x^2 - then I lose it because of the canceling out, and I'm unsure what happens to the x's...

    I know the answer is -1/6 (2x^3 - 3) + C, but would appreciate the working I miss out. Or if I am attempting it using the wrong methods.

    Many thanks,

    Ocis
     
  2. jcsd
  3. May 24, 2008 #2
    you get [tex]\int[/tex]1/6u^2 du, so now integrate 1/u^2 with respect to u
     
  4. May 24, 2008 #3
    Thank you for the reply.

    Ok so can I say that 1/u^2 = u^-2 ?
    if so it would become 1/6 . (u^-1 / -1) = -1/6 (u) ??? = -1/6 (2x^3 - 3) + C

    Ok if that is right I think I understand a little more but just get confused with what to do with the x^2 as the numerator...?
     
  5. May 24, 2008 #4
    dx=du/6[tex]x^{2}[/tex]
    when you substitute this in the expression the x^2 in the numerator cancels out, so there is in fact no x left in the expression.
     
  6. May 24, 2008 #5
    Ok I understand now, thank you very much.
     
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