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Quick Question

  1. Sep 10, 2005 #1
    [tex] \int \frac {1}{x\sqrt{4x+1}}dx[/tex]

    Here's what I have done so far on this problem

    I let [tex] u= \sqrt{4x+1} [/tex], so then [tex] u^2=4x+1 [/tex], [tex] du= \frac {2dx}{u} [/tex] and [tex] x= \frac {u^2-1}{4} [/tex]

    Substituting, I get [tex] \int \frac {1}{(\frac{u^2-1}{4})u}du [/tex]

    Then moving stuff around, I get [tex] 4 \int \frac {du}{u(u+1)(u-1)} [/tex]

    I know I have to use partial fractions. But I am not sure where to start. Any help is appreciated.
  2. jcsd
  3. Sep 10, 2005 #2
  4. Sep 10, 2005 #3
    I know that, but then what do I do, I have to get a common denominator, which is [tex] u(u+1)(u-1) [/tex], but then what?
  5. Sep 10, 2005 #4
    Multiply both sides by u(u+1)(u-1) so that the left side is 1. Simplify to get A, B, and C.
  6. Sep 10, 2005 #5
    Ok, I get A=-1, b=1/2, c=1/2, is that right?
  7. Sep 10, 2005 #6
    Looks good.
  8. Sep 10, 2005 #7
    Ok, I think I got it. Thanks.
  9. Sep 11, 2005 #8


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    Homework Helper

    Just a little bit missed out. When substituting into the integrand written in terms of x, you forgot to replace the dx with (u/2)du.

    [tex]\mbox{let}\ u= \sqrt{4x+1}\ \mbox{, so then}\ u^2=4x+1\ \mbox{,}\ du= \frac {2dx}{u}\ \rightarrow dx = \frac{u}{2}\ du \mbox{, and}\ x= \frac {u^2-1}{4} [/tex]

    Substituting gives [tex]\int \frac {1}{x\sqrt{4x+1}}dx = \int \frac {1}{(\frac{u^2-1}{4})u}.\frac{u}{2}\ du = \int \frac{2}{u^2 - 1}\ du[/tex]

    Now you can do the partial fractions bit.
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