Results 1 to 13 of 13

Math Help - [SOLVED] Integration and natural logs

  1. #1
    Junior Member Coco87's Avatar
    Joined
    Jan 2008
    Posts
    45
    Awards
    1

    [SOLVED] Integration and natural logs

    I have run into a brick wall that is driving me insane (see following picture: )

    It seems simple, yet I'm not getting the answer in the book, but rather I'm looping through the Integration By Parts method over and over again.

    \int{\ln{(2x+1)}dx}

    I set:

    u=\ln{(2x+1)} dv=dx
    du=\frac{2}{2x+1} v=x

    and get:

    x\ln{(2x+1)}-\int{\frac{2x}{2x+1}}

    and from there I integrate by partial fractions:

    2\int{x\frac{1}{2x+1}}

    u=\frac{1}{2x+1} dv=x
    du=\frac{2}{(2x+1)^2} v=\frac{x^2}{2}

    then I get:

    2[\frac{x^2}{2(2x+1)}-\int{\frac{x^2}{(2x+1)^2}}]

    I might be wrong, but it seems like an endless loop to me

    Could anyone help me out?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    If you integrate by parts once, then integrate by parts with the reverse substitution, it is sure you won't get to the answer

    After the first integration by parts, you have :

    \int \frac{2x}{2x+1} ~dx

    =\int \frac{(2x+1)-1}{2x+1} ~dx=\int 1-\frac{1}{2x+1} ~dx=\int ~dx-\int \frac{1}{2x+1} ~dx

    Can you do it ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Use the sub: u = 2x + 1 \ \Rightarrow \ du = 2 dx \iff \frac{du}{2} = dx

    So: \int \ln (2x + 1) \ dx \ = \ \frac{1}{2} \int \ln u \ du

    which is a standard one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    792
    \int{\frac{2x}{2x+1}}
    Why use integration by parts for this one? Simply add and subtract one to get:
    \int (1 - \frac{1}{2x+1})

    It should be simple enough now.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by o_O View Post
    which is a standard one.
    Standard one that you get with an integration by parts. It may not sound as standard as you know it
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member Coco87's Avatar
    Joined
    Jan 2008
    Posts
    45
    Awards
    1
    Quote Originally Posted by Moo View Post
    Hello,

    If you integrate by parts once, then integrate by parts with the reverse substitution, it is sure you won't get to the answer

    After the first integration by parts, you have :

    \int \frac{2x}{2x+1} ~dx

    =\int \frac{(2x+1)-1}{2x+1} ~dx=\int 1-\frac{1}{2x+1} ~dx=\int ~dx-\int \frac{1}{2x+1} ~dx

    Can you do it ?
    Hmm, where did you get the -1 from on the first part?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Coco87 View Post
    Hmm, where did you get the -1 from on the first part?
    1-1=0

    I made "appear" the denominator in order to simplify. It is a similar process to a polynomial division.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member Coco87's Avatar
    Joined
    Jan 2008
    Posts
    45
    Awards
    1
    Quote Originally Posted by Moo View Post
    1-1=0

    I made "appear" the denominator in order to simplify. It is a similar process to a polynomial division.
    Ah, thanks! , I didn't know that is allowed.

    I'm still having an issue though

    The answer I get is:

    (x-\frac{1}{2})\ln{(2x+1)}-x+C

    However, the answer in the back of the book is:

    \frac{1}{2}(2x+1)\ln{(2x+1)}-x+C

    I'm not sure where I messed up
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Coco87 View Post
    Ah, thanks! , I didn't know that is allowed.

    I'm still having an issue though

    The answer I get is:

    (x-\frac{1}{2})\ln{(2x+1)}-x+C

    However, the answer in the back of the book is:

    \frac{1}{2}(2x+1)\ln{(2x+1)}-x+C

    I'm not sure where I messed up


    \frac 12 (2x-1)=\frac 12 (2x)-\frac 12=\dots ??



    and by the way, it must be the absolute value of 2x+1 in the logarithm, since it is not defined over non-positive numbers.
    Last edited by Moo; September 23rd 2008 at 11:31 AM. Reason: oooh it's a minus sign ! but it doesn't change much :D
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    When you did integration by parts you subtract the integral ( -\int vdu. You broke up that integral into two parts. Did you distribute the negative sign to both?

    After that change of sign, they just brought out a constant to make it look neater.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member Coco87's Avatar
    Joined
    Jan 2008
    Posts
    45
    Awards
    1
    I see how they connect now, I'm still unsure where they got the 2x from to multiply by
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    (x+\frac{1}{2}) = \frac{1}{2}(2x+1)

    ...
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member Coco87's Avatar
    Joined
    Jan 2008
    Posts
    45
    Awards
    1
    Ah, now I got ya! I was looking at it the wrong way I guess (long day).

    Thanks for helping me out guys!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 05:39 PM
  2. Integration with Natural Logs
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 8th 2010, 02:45 PM
  3. [SOLVED] Natural Logarithmic Integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 9th 2009, 03:19 PM
  4. Replies: 4
    Last Post: August 31st 2008, 06:26 PM
  5. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 06:18 AM

Search Tags


/mathhelpforum @mathhelpforum