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Calc II Integral

  1. Jun 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve.

    [tex]
    \int \frac{xdx}{\sqrt{x^2 + 4x}}
    [/tex]


    3. The attempt at a solution

    First I added and subtracted 2, then let [itex] u = x^2 + 4x [/itex]

    After that I fiddled around with it to no success. What's my problem?

    [tex]
    \int \frac{xdx}{\sqrt{x^2 + 4x}} \Rightarrow \int \frac{(x+2)dx}{\sqrt{x^2 + 4x}} - \int \frac{2dx}{\sqrt{x^2 + 4x}} \Rightarrow
    \int \frac{u^{-1/2}}{2} - \int \frac{2dx}{\sqrt{x^2 + 4x}} = (x^2 + 4x)^{1/2} - \int \frac{2dx}{\sqrt{x^2 + 4x}}
    [/tex]


    Thanks.

    edit: got it thanks guys
     
    Last edited: Jun 22, 2010
  2. jcsd
  3. Jun 21, 2010 #2
    I would try trig substitution. Have you learned it?
     
  4. Jun 21, 2010 #3
    oh. the 4x was throwing me off, but I guess I can just make it [itex] (\sqrt{x})^2 [/itex].

    Thanks.
     
  5. Jun 21, 2010 #4

    Mark44

    Staff: Mentor

    x2 + 4x = x2 + 4x + 4 - 4 = (x + 2)2 - 4

    Now you're ready for a substitution.
     
  6. Jun 22, 2010 #5

    Gib Z

    User Avatar
    Homework Helper

    Maybe it was his trap. If the OP were to use his solution he would have to write and still do much much more than if he listened to Mark44 and Tedjn. And his teacher might mention in his feedback how unnecessary that was.
     
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