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How to find the integral of x(3x^2 + 5)^1/4

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data

    2. The attempt at a solution
    u=3x^2+5 du=6x

    this is where i'm getting sloppy, i think...


    and then i get stuck.
  2. jcsd
  3. Apr 28, 2008 #2
    Good so far...(except du=6xdx and you forgot to include a few dummy variables throughout the problem, but you are accounting for them in your attempt--just remember to write them down too)

    Where are you getting the integral sign from in this part of the problem?
  4. Apr 28, 2008 #3
    i was trying to use the power rule for integrals (x^n+1/n+1). but i have a feeling i can't do that. haha :)

    then i pulled the 1/6 back into it and multiplied it out (1/6 times 4/5=2/15).
  5. Apr 28, 2008 #4
    No, you were doing it right, no need to second guess yourself. The power rule for integrals is all you need to solve this.

    Try to integrate the following again:


    Rather, another way for me to put it:

    is [tex]\int[/tex]udu = [tex]\int[/tex]0.5u^2du ?
    Last edited: Apr 28, 2008
  6. Apr 28, 2008 #5
    no, they wouldn't be equal. for example, does 5=0.5(5)^2? no; 5 does not equal 12.5

    did i pull my denominator up incorrectly? if (x^n+1/n+1) is (x^.25+1/.25+1), then (x^1.25/1.25). so to pull up the denominator...
  7. Apr 28, 2008 #6
    Ok, if those two aren't equal, then why would

    1/6[tex]\int[/tex]du(u)^1/4 = 1/6[tex]\int[/tex]4/5(u)^5/4

    Your denominator and everything is fine, you just have an extra integral sign floating around after you already correctly evaluated the integral. I didn't outright say it since I thought you'd notice it, but I suppose my references were a bit too vague.

    Once you get rid of the extra integral sign at the point where you are "stuck", you would have 2/15(3x^2+5)^5/4 (+C), and that would simply be your final answer, unless there was an initial condition or something.
  8. Apr 28, 2008 #7
    oh wow, that's what i get for trying to do calculus late at night. i was trying to evaluate an integral after i already evaluated it. so my final answer was correct... just need to watch out and make sure i drop the integral sign.

    should have been...

    thank you so much.
    Last edited: Apr 28, 2008
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