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Antiderivative and Indefinite Integration

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    ∫(x3 - 3x2 + x + 1)/√x dx

    3. The attempt at a solution

    ∫x3-1/2 - 3∫x2-1/2 + ∫x1-1/2 + ∫x1-1/2

    ∫x5/2 - 3∫x3/2 + ∫x1/2 + ∫1/2

    (x7/2)7/2 - (3x5/2)5/2 + (x3/2)3/2 + (x3/2)3/2 + C

    (2x7/2)/7 - (6x5/2)/5 + 2x3/2)/3 + (2x3/2)/3 + C

    Thank you so much!!
     
  2. jcsd
  3. Mar 5, 2013 #2
    First, the last integral you wrote, should be [tex]∫{\frac{1}{√x}}[/tex]
    Second, When integrating, you divide by the new power, you are multiplying, so you should 2/7, 2/5.. etc..

    Edit: Im not sure what you did on the line jsut after integrating, but the line after it is mostly right, except for that last part.
     
  4. Mar 5, 2013 #3
    I don't know if I understand it.

    If I have:

    ∫x3/√x dx

    = ∫x3 - 1/2
    = ∫x5/2
    = (x5/2 + 1)/ 5/2 +1
    = (x7/2)/7/2 + C

    Simplyfying:

    2x7/2/7 + C

    Is this correct?

    If not, what am I doing wrong?

    Also, I am not sure how to proceed with ∫1/√x dx

    If I have ∫1 dx:

    = 1 + C

    So for ∫1/√x dx I'll have:

    =x/√x
    =x . x- 1/2
    =x1/2
    =(x1/2+ 1)/1/2 +1 + C
     
  5. Mar 5, 2013 #4
    The first part was right. for ∫1/√x dx. rewrite it as ∫x-1/2dx

    EDIT: Nowhere in the equation would you have ∫1dx, but if you did. it would be x+c, not 1+c.
     
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