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Possibly a Simple integration?

  1. May 29, 2008 #1
    This integration might be very simple but I feel like I'm missing something here. I tried integration by parts but I ended up with a mess, x on top of x and the integral never came out without two x's as a product of each other.


    Any help is appreciated, I'm trying to learn Calculus myself.
    Last edited: May 29, 2008
  2. jcsd
  3. May 29, 2008 #2
    Hint: what is d/dx(ln(x))?
  4. May 29, 2008 #3
    let [tex] 4x=y [/tex]


    [tex]\int\frac{ln 4x}{2x}dx = \frac{1}{2}\int\frac{lny}{y} dy[/tex]

    integrate by parts :

    [tex] lny = u , dv=\frac{dy}{y} [/tex]


    [tex] du=\frac{dy}{y} , v=lny [/tex]


    [tex]\int\frac{ln y}{y} dy = {(lny)}^2 - \int\frac{ln y}{y} dy [/tex]


    [tex]\int\frac{ln y}{y} dy = \frac{1}{2} {(lny)}^2 [/tex]


    [tex]\int\frac{ln 4x}{2x}dx = \frac{1}{4} {(ln4x)}^2 +C [/tex]
    Last edited: May 29, 2008
  5. May 29, 2008 #4

    Gib Z

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    Homework Helper

    No, do NOT integrate by parts. Use maze's hint.
  6. May 29, 2008 #5
    ok , maze's hint is smart , but i thought integrating by parts would be more informative as metalmagik said that he is learning !!
  7. May 29, 2008 #6
    Whoa, I didn't even know anyone replied to this yet.

    Well mmzaj's strategy looks like it works, but d/dx of ln(x) is 1/x...

    if I use u substitution for this problem, I get u = ln(4x), du = 1/x dx

    So, I have to put a 2 outside the integral to balance.

    2[tex]\int[/tex]u du
    2 [tex]\frac{u^2}{2}[/tex]

    Then the 2's cancel out and I'm left with u^2 or ln(4x)^2. But this is not what mmzaj got from integration by parts.

    Unless my u-substitution is off...

    Why is there a problem?
  8. May 29, 2008 #7


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    f(x)=ln(u) f'(x)=u'/u
  9. May 30, 2008 #8
    The 2 should be in the denominator when you pull it out, and then multiplying by the 1/2 from u^2/2, that gets the factor of 1/4.
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