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Use a substitution to compute the integral

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Going over past exam problems, stuck on this one. Attached


    Calc 2, topics include for this exam integration techniques, such as partial fractions, improper integrals, trig sub, and series.

    Question reads: Use a substitution to compute: (see attached)


    2. Relevant equations



    3. The attempt at a solution

    I tried partial fractions integration.

    1. dx/x(1 + sqrt(x)

    2. A/x + B/(1+sqrt(x) = 1/x(1+sqrt(x))

    Solved for A, A= 1 B = -1


    Resulted in:

    1/x - 1/1+sqrt(x)

    I integrated each part:


    ln(abs(x)) - (ln(abs(1+sqrt(x)))*(1/2sqrt(x))


    I am not sure if this was correct.
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2014 #2

    ehild

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    [itex]\frac{1}{x(1+\sqrt{x}) }[/itex] is not equal to 1/x - 1/(1+sqrt(x)).

    Use the substitution √x = u first.


    ehild
     
  4. Mar 11, 2014 #3
    u = sqrt(x)
    du = 1/2sqrt(x)

    x = u^2

    ∫1/(u^2 + u^3)

    ∫1/(u^2(1+u))

    Then I use partial fractions decomposition

    A= 1
    B = -1

    ∫1/u^2 - ∫1/(1+u)

    substitute in u

    ∫1/x - ∫1/(1+sqrt(x))


    ln(abs(x)) - ln(abs(1+sqrt(x)))

    Answer: ln((x/(1+sqrt(x)))
     
  5. Mar 11, 2014 #4

    ehild

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    That is wrong...

    Determine dx in terms of u and du. Without dx or du, it is not an integral!

    ehild
     
  6. Mar 11, 2014 #5

    Simon Bridge

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    Did you forget to sustitute for dx?

    Make it easier for you: Define ##x=u^2 \implies dx=?##
     
  7. Mar 11, 2014 #6
    so how does this sound:


    2u/(u^2(1+u)) du

    Then I complete the partial fraction decomposition?
     
  8. Mar 11, 2014 #7

    Curious3141

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    Simplify by cancelling first, then do the partial fraction decomposition.
     
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