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Integration question, u substitution

  1. Feb 8, 2013 #1
    The problem statement
    Evaluate the indefinite integral

    ∫[itex]\frac{\sqrt{x}}{\sqrt{x}-3}[/itex]dx

    The attempt at a solution
    My first thought was to substitute u for √(x)-3, but then du would equal [itex]\frac{1}{2\sqrt{x}}[/itex]dx, and there's no multiple of du in the integrand.

    Next, I tried splitting up the problem like this:
    ∫[itex]\frac{1}{\sqrt{x}-3}[/itex]*[itex]\sqrt{x}[/itex]
    When u is substituted for the [itex]\sqrt{x}[/itex], du still equals[itex]\frac{1}{2\sqrt{x}}[/itex]dx , but the minus 3 in the denominator of the first factor prevents cleanly substituting a multiple of du for [itex]\frac{1}{\sqrt{x}-3}[/itex]dx.

    Any ideas or hints?
     
  2. jcsd
  3. Feb 8, 2013 #2
    Notice that if ##u = \sqrt{x} - 3##, then ##u + 3 = \sqrt{x}##.

    Since ##\displaystyle du = \frac{1}{2\sqrt{x}} dx, 2\sqrt{x}\ du = dx##

    Does this make things easier?
     
  4. Feb 8, 2013 #3

    HallsofIvy

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    [itex]\sqrt{x}dx= (x/\sqrt{x})dx= (\sqrt(x)^2)(dx/\sqrt{x})= (u+3)^2du[/itex]
     
  5. Feb 8, 2013 #4
    The denominator is kind of hard to see, it's sqrt(x)-3. Thanks for all the replies, I'll see if I can figure it out
     
    Last edited: Feb 8, 2013
  6. Feb 8, 2013 #5

    SammyS

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    If [itex]\displaystyle \ \ du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx\,,[/itex]

    then [itex]\displaystyle \ \ dx=2u\,du\ .[/itex]

     
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