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Integration question

  1. Oct 2, 2005 #1
    [tex]\int \frac{1}{x\sqrt{4x + 1}}dx[/tex]
    I let [tex]u= \sqrt{4x+1}[/tex], [tex]u^2 = 4x+1[/tex]
    So [tex]\frac{1}{2}u du = dx[/tex]

    [tex]\int \frac{\frac{1}{2}u}{(\frac{u^2-1}{4})*((\sqrt{u})^2)}du[/tex]
    [tex]\int \frac{2u}{(u^2 - 1)*(u)} du[/tex]
    [tex]\int \frac{2}{u^2 - 1} du[/tex]
    [tex]2\int \frac{1}{u^2 -1}du[/tex]

    Can anyone help me anti-differentiate what the integrand is?
     
    Last edited: Oct 2, 2005
  2. jcsd
  3. Oct 2, 2005 #2

    James R

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    Isn't it related to the inverse hyperbolic tangent of u?
     
  4. Oct 2, 2005 #3
    Try u=sinθ.
     
  5. Oct 3, 2005 #4
    I'd suggest partial fraction decomposition...
     
  6. Oct 3, 2005 #5

    TD

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    Expanding that fractions gives:

    [tex]\frac{1}
    {{u^2 - 1}} = \frac{1}
    {{2\left( {u - 1} \right)}} - \frac{1}
    {{2\left( {u + 1} \right)}}[/tex]
     
  7. Oct 3, 2005 #6

    James R

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    Partial fractions work. So does u=tanh t.
     
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