Results 1 to 8 of 8

Math Help - lnx Integral

  1. #1
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106

    lnx Integral

    \int \frac{ln|x|} {x\sqrt{1+ln|x|}} ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let t=\ln x ---> trivial excercise.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    I don't get it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    If t=\ln x\implies dt=\frac1x\,dx.

    It's just substitution.

    Can you handle it from there?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    but then I get

    \int\frac{u}{\sqrt{1+u}}~dt

    how do I integrate this?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    Substitute \varphi ^2 = 1 + u,

    \int {\frac{u}<br />
{{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member akhayoon's Avatar
    Joined
    Dec 2007
    From
    T.O
    Posts
    106
    Quote Originally Posted by Krizalid View Post
    Substitute \varphi ^2 = 1 + u,

    \int {\frac{u}<br />
{{\sqrt {1 + u} }}\,du} = 2\int {\left( {\varphi ^2 - 1} \right)\,d\varphi } .
     \int\frac{\varphi^{2}-1}{\varphi}~d\varphi

    so to get ur answer we had to turn \frac{1}{\varphi} into 2?

    how does that work

    I'm sorry but I'm just trying to understand the solution
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by akhayoon View Post
     \int\frac{\varphi^{2}-1}{\varphi}~d\varphi

    so to get ur answer we had to turn \frac{1}{\varphi} into 2?

    how does that work

    I'm sorry but I'm just trying to understand the solution
    Krizalid thinks a few levels above the average person when it comes to integrals. here is a more conventional substitution for mortals.

    \int \frac u{\sqrt{1 + u}}~du

    Let t = 1 + u \implies \boxed{u = t - 1}

    \Rightarrow dt = du

    So our integral becomes:

    \int \frac {t - 1}{\sqrt{t}}~dt

    which we can do using the power rule (after dividing \sqrt{t} into each of the terms in the numerator)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum