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Can you integrate?

  1. Aug 12, 2009 #1
    Integrate the following--->
    {x3+1/(whole root over)x2+x}dx
     
  2. jcsd
  3. Aug 12, 2009 #2
    Do you mean [tex]\int \frac{x^3+1}{\sqrt{x^2+x}}\,dx[/tex] or [tex]\int \left(x^3+\frac{1}{\sqrt{x^2+x}}\right)dx?[/tex]

    Either way, the best thing to do is to start by completing the square in the denominator, then using a bunch of trig substitution stuff.
     
  4. Aug 12, 2009 #3
    x3+13=(x+1)(x2-x+1)
    x2+x=x(x+1)

    Is it enough help?
     
    Last edited: Aug 13, 2009
  5. Aug 12, 2009 #4

    Mark44

    Staff: Mentor

    Actually, x3 + 1 = (x + 1)(x2 - x + 1).

    In any case, we still don't know exactly what the integrand is.
     
  6. Aug 13, 2009 #5
    I mean the first image.
     
  7. Aug 13, 2009 #6
    Mark44 thanks for the correction.

    perfectibilis start by writing x3+1 with

    [tex]\sqrt{(x+1)^2(x^2-x+1)^2}[/tex]
     
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