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Integral Help

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data
    As I was reviewing some of my previuosly learned calculus I came across somthing that I had either forgoten how to do or was never taught. How do you take the integral of somthing like this

    [tex]\int \sqrt{\frac{9}{4}x+1}[/tex]

    I don't know how to start with this one. Do I use U substitution?
     
  2. jcsd
  3. Aug 18, 2009 #2
    First rewrite the integral as

    dy=[tex]\int(\frac{9}{4x}+1)^{1/2}[/tex] dx

    Now use the u substitution.

    Thanks
    Matt
     
  4. Aug 18, 2009 #3
    Using U substitution I got ((9/4x+1)^1/2)/(27/8) I don't think I did this right.
     
  5. Aug 18, 2009 #4
    I don't believe that is correct.

    Try this.

    Let u = [tex]\frac{9}{4x}[/tex]+1

    Now find du/dx.
     
  6. Aug 18, 2009 #5
    Du/dx is =36/16x[tex]^{2}[/tex]
     
  7. Aug 18, 2009 #6
    Can you show your steps as to how you got that result please?
     
  8. Aug 18, 2009 #7
    u=[tex]\frac{9}{4x}+1 [/tex]

    [tex]\frac{0*4x-9*4}{(4x)^{2}}[/tex] So I made a mistake with the negative? The answer should be -36/16x^2.
     
  9. Aug 18, 2009 #8
    Yes, you got it. Now perform the rest of the u substitution procedure and your done.
     
  10. Aug 18, 2009 #9
    I think the discussion got off track somewhere. The radicand above is [itex] \frac{9}{4}x+1[/itex] not [itex]\frac{9}{4x}+1[/itex]. Or was the original statement an error?

    If the original is correct, then it can be easilly integrated by substitution - try [itex]u=\frac{9}{4}x+1[/itex].

    (This looks like a Stewart's Calculus exercise.)

    --Elucidus
     
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