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Finding the integral of a function

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi, all I need to do is find the integral of a function, and im struggling with where to begin with this particular expression.


    2. Relevant equations

    [itex]\int[(3x^2)/(x^3+ 2)]dx[/itex]

    3. The attempt at a solution

    Thats the problem, im not sure where to start! Im sure it is something obvious I am missing.
    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 24, 2011
  2. jcsd
  3. Oct 24, 2011 #2

    HallsofIvy

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    I assume you mean [itex]\int[(3x^2)/(x^3+ 2)]dx[/itex].

    Let [itex]u= x^3+ 2[/itex].
     
  4. Oct 24, 2011 #3
    Thanks for your quick reply. Sorry, that is exactly what I meant, Im not great with laying stuff out online.

    Substituting was my first thought, however this question is part of a pre-test Im doing for homework, and there is a section on substituting later on in the paper, so Im guessing I'm not supposed to use that here. Is there another way doing it?
     
  5. Oct 24, 2011 #4
    For equations of this form, where the numerator is the derivative of the denominator, the integral is simply the natural logarithm of the denominator. So in this case it would be ln|x^3+2| + a constant of integration.
     
  6. Oct 24, 2011 #5

    Mark44

    Staff: Mentor

    Substitution is probably the easiest technique, and the one that should be tried first. I can't think of any other way to do this problem.
     
  7. Oct 24, 2011 #6
    I never knew that, that works every time? Care to explain how?
     
  8. Oct 24, 2011 #7

    HallsofIvy

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    Because of "substitution" whether you call it that or not! If your integral is of the form [itex]\int f'(x)/f(x)dx[/itex], let u= f(x), du= f'dx changes the integral to [itex]\int du/u[/itex].
     
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