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Easy u sub

  1. Feb 29, 2012 #1
    question: integral of [(3-ln2x)^3]/2x

    my workings:

    I let u = 3-ln2x
    then du= -2/x dx
    so -1/2du = 1/x dx

    this leaves me with -(1/2)*integral of u^3/2 du

    I take the bottom 2 out to get -(1/2)*(1/2) * integral of u^3 du

    which is -1/4 * (u^4)/4

    then I sub u in to get

    -1/4 *(3-ln2x)/4

    Which is -1/16 * (3-ln2x)

    On my test I only got 8/10 for this question and When I plug this example into Wolfram it gives some wacky answer.

    Can someone tell me if I'm right? Thanks
  2. jcsd
  3. Feb 29, 2012 #2


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    Gold Member

    No. du = -(1/x)dx. You need the chain rule there.

    You left off the 4th power. Fix those two mistakes and it is correct.
  4. Feb 29, 2012 #3
    Oops.. the 4th power I did write on my test just missed it there..

    and damn I cant believe I got that derivative wrong haha

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