1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?