Results 1 to 7 of 7

Math Help - integrating ln

  1. #1
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39

    integrating ln

    I'm trying to integrate \int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx
    I start with
    u = \ln(x+\sqrt{1+x^2})
    \frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}}  )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}

    Can I do like this? \ln{x} + \ln{\sqrt{1+x^2}}=\ln(x+\sqrt{1+x^2})
    if so, then do I continue like this? \int{2u} du= u^2


    Hm... and I noticed that I rewrited \ln backward xD omg, can anyone help me out here?
    Last edited by Revy; August 24th 2010 at 02:20 PM. Reason: noticed smth
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    A clue here is the \sqrt{1 + x^2} ... if that x were a tan of something, how would that change things?

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case theta), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    The general drift is...



    Spoiler:


    Late edit: Soroban's solution below is a bit simpler, as it does without the trigonometry. Just in case a picture helps...



    Spoiler:


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; August 25th 2010 at 08:04 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Revy View Post
    I'm trying to integrate \int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx
    I start with
    u = \ln(x+\sqrt{1+x^2})
    \frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}}  )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}

    Can I do like this? \ln{x} + \ln{\sqrt{1+x^2}}=\ln(x+\sqrt{1+x^2})
    if so, then do I continue like this? \int{2u} du= u^2


    Hm... and I noticed that I rewrited \ln backward xD omg, can anyone help me out here?
    No, the logarithm of a sum is not the sum of the logarithms! You should review the basic properties of logarithms.

    Here's a hint : the numerator looks very much like the expressions one gets for the inverse hyperbolic functions (such as arcsinh, arctanh, etc).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    630
    Hello, Revy!

    I got the answer . . . but it was a strange journey.


    \displaystyle{\int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt  {1+x^2}}\,dx

    We have: . \displaystyle \int \ln(x+\sqrt{1+x^2})\cdot\frac{dx}{\sqrt{1+x^2}} .[1]

    I used the same substitution, but I found a sneaky step.


    Let u \,=\,x + \sqrt{1+x^2} .[2]

    Then: . du \:=\:\left(1 + \dfrac{x}{\sqrt{1+x^2}}\right)\,dx \quad\Rightarrow\quad  du\:=\:\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\,dx

    From [2], we have: . du \:=\:\dfrac{u}{\sqrt{1+x^2}}\,dx \quad\Rightarrow\quad \dfrac{dx}{\sqrt{1+x^2}} \,=\,\dfrac{du}{u} .[3]


    Substitute [2] and [3] into [1]: . \displaystyle \int \ln u\cdot\frac{du}{u}

    Let v \,=\,\ln u \quad\Rightarrow\quad dv \,=\,\frac{du}{u}

    Substitute: . \displaystyle{\int v\,dv \;=\;\tfrac{1}{2}v^2 + C


    Back-substitute: . \frac{1}{2}[\ln u]^2 + C

    Back-substitute: . \frac{1}{2}\bigg[\ln\!\left(x+\sqrt{1+x^2}\,\right)\bigg]^2 + C

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Ted
    Ted is offline
    Member
    Joined
    Feb 2010
    From
    China
    Posts
    199
    Thanks
    1
    The substitution u=ln\left(x+\sqrt{1+x^2}\right) will solve it directly.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39
    Could you explain me this part? \dfrac{dx}{\sqrt{1+x^2}} \,=\,\dfrac{du}{u}

    According to book answer should be \ln^2\!\left(x+\sqrt{1+x^2}\,\right) + C<br />
but I see that more than one person got the same answer, then there might be mistake in the book


    Quote Originally Posted by Ted View Post
    The substitution u=ln\left(x+\sqrt{1+x^2}\right) will solve it directly.
    If I substitute whole ln then it goes like that?
    du=\frac{1}{x+\sqrt{1+x^2}} \frac{x+\sqrt{1+x^2}{\sqrt{1+x^2}}}dx=\frac{dx}{\s  qrt{1+x^2}}
    and then as was posted before
    \int {\ln{u}du}
    v=\ln u \quad\Rightarrow\quad dv =\frac{du}{u}
    \displaystyle{\int v\,dv \;=\;\tfrac{1}{2}v^2 + C
    Last edited by Revy; August 25th 2010 at 05:41 AM. Reason: for not double posting
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Revy View Post
    I'm trying to integrate \int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx

    I start with

    u = \ln(x+\sqrt{1+x^2})

    correct

    \frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}}  )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}

    No! the \displaystyle\;\;\frac{1}{x}\;\; term should be \diplaystyle\;\;\frac{1}{x+\sqrt{1+x^2}}

    Think of it as

    u=lny\Rightarrow\ \frac{du}{dx}=\frac{du}{dy}\ \frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}

    Can I do like this?

    \ln{x} + \ln{\sqrt{1+x^2}}=\ln(x+\sqrt{1+x^2})

    This breaks the laws of logarithms.


    if so, then do I continue like this? \int{2u} du= u^2


    Hm... and I noticed that I rewrited \ln backward xD omg, can anyone help me out here?

    u=ln\left(x+\sqrt{1+x^2}\right)=lny

    \displaystyle\frac{du}{dx}=\frac{1}{y}\ \frac{dy}{dx}=\frac{1}{x+\sqrt{1+x^2}}\left(1+\fra  c{2x}{2\sqrt{1+x^2}}\right)

    =\displaystyle\frac{1}{x+\sqrt{1+x^2}}\left(1+\fra  c{x}{\sqrt{1+x^2}}\right)=\frac{1}{x+\sqrt{1+x^2}}  \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right)

    =\displaystyle\frac{1}{\sqrt{1+x^2}}

    \Rightarrow\int{u}du=\frac{u^2}{2}+C

    since...

    \displaystyle\frac{du}{dx}=\frac{1}{\sqrt{1+x^2}}\  Rightarrow\ du=\frac{dx}{\sqrt{1+x^2}}\Rightarrow\int{ln\left(  x+\sqrt{1+x^2}\right)\left(\frac{dx}{\sqrt{1+x^2}}  \right)=\int{u}du
    Last edited by Archie Meade; August 25th 2010 at 06:24 AM. Reason: additional explanation to answer added question
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need help on integrating this
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 4th 2010, 06:29 AM
  2. Integrating
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 2nd 2010, 11:57 AM
  3. integrating
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 31st 2010, 02:18 PM
  4. integrating e
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 27th 2010, 11:03 PM
  5. Integrating (2x^2 -3)/x(x^2 +4
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 17th 2010, 06:14 AM

/mathhelpforum @mathhelpforum