Results 1 to 6 of 6

Thread: integration by part

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    18

    integration by part

    I'm having some trouble with this

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

    I set $\displaystyle u = ln(2x+1)$ and $\displaystyle dv = dx$ so that $\displaystyle du=\frac{2}{2x+1}$ and $\displaystyle v = x$.

    so after integrating by part once I get:

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

    then I tried to integrate by part again with $\displaystyle u=x$ and $\displaystyle dv=(2x+1)^{-1}$

    my final answer ends up as $\displaystyle x\ln{(2x+1)}$ but my solution manual tells me differently.

    Can anyone tell me what I'm doing wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by stevedave View Post
    I'm having some trouble with this

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

    I set $\displaystyle u = ln(2x+1)$ and $\displaystyle dv = dx$ so that $\displaystyle du=\frac{2}{2x+1}$ and $\displaystyle v = x$.

    so after integrating by part once I get:

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

    then I tried to integrate by part again with $\displaystyle u=x$ and $\displaystyle dv=(2x+1)^{-1}$

    my final answer ends up as $\displaystyle x\ln{(2x+1)}$ but my solution manual tells me differently.

    Can anyone tell me what I'm doing wrong?
    For the integral $\displaystyle -\int\frac{2x}{2x+1}\,dx$, make the substitution $\displaystyle z=2x+1\implies x=\tfrac{1}{2}\left(z-1\right)$

    Thus, $\displaystyle \,dz=2\,dx$.

    The integral becomes: $\displaystyle -\int\frac{x\,dz}{z}\implies -\tfrac{1}{2}\int \frac{z-1}{z}\,dz=\tfrac{1}{2}\int\left(\frac{1}{z}-1\right)\,dz$

    Can you continue from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Quote Originally Posted by stevedave View Post
    I'm having some trouble with this

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

    I set $\displaystyle u = ln(2x+1)$ and $\displaystyle dv = dx$ so that $\displaystyle du=\frac{2}{2x+1}$ and $\displaystyle v = x$.

    so after integrating by part once I get:

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

    then I tried to integrate by part again with $\displaystyle u=x$ and $\displaystyle dv=(2x+1)^{-1}$

    my final answer ends up as $\displaystyle x\ln{(2x+1)}$ but my solution manual tells me differently.

    Can anyone tell me what I'm doing wrong?
    In the second integral, expand

    $\displaystyle \int\frac{x}{2x+1}dx = \int \frac{1}{2} - \frac{1}{2}\frac{1}{2x+1} dx$ and integrate each term separately

    BTW $\displaystyle \int \frac{1}{ax+b} dx = \frac{1}{a} \ln| ax + b| + c$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    $\displaystyle \int \ln(2x+1) \, dx = x\ln(2x+1) - \int \frac{2x}{2x+1} \, dx
    $

    note that $\displaystyle \frac{2x}{2x+1} = \frac{2x+1-1}{2x+1} = 1 - \frac{1}{2x+1}$

    $\displaystyle \int \ln(2x+1) \, dx = x\ln(2x+1) - \int 1 - \frac{1}{2x+1} \, dx$

    can you finish from here?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2009
    Posts
    18
    Thank you everyone. That was very helpful.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    How about this for a short cut? Notice that one integral of 1 is x + 1/2, and make that the value of v, for dv/dx = 1 ...





    As usual, straight lines differentiate down / integrate up with respect to x.

    Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration by part
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 4th 2011, 07:36 AM
  2. twice integration by part?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 24th 2011, 08:03 PM
  3. Integration with a 2 part du
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 17th 2009, 03:34 AM
  4. Integration by Part
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Nov 8th 2008, 01:51 PM
  5. integration part 2
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Feb 26th 2008, 11:20 AM

Search Tags


/mathhelpforum @mathhelpforum