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Integration by Parts of Inverse Tangent

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data
    I must evaluate the indefinite integral:
    [tex]\int x \arctan{x} dx[/tex]


    2. Relevant equations
    I am using the following format to perform the integration:
    [tex]\int u dv = uv - \int v du[/tex]

    3. The attempt at a solution
    I have tried working the problem substituting x in for u and arctan in for dv and here is my work:

    [tex] Let \ u=x; \ du = dx[/tex]
    [tex]Let \ v = \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}; \ dv = \arctan{x}[/tex]
    [tex]\int x \arctan{x} dx = x (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) - \int (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) dx[/tex]
    [tex]2 \int \arctan{x} dx = x (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) + \int \frac{1}{2} \ln{(1 + x^2)} dx[/tex]

    I then apply another integration by parts technique:

    [tex]Let \ u = \frac{1}{2} \ln{(1 + x^2)}; \ Let \ du = \frac{x}{1 + x^2} dx[/tex]
    [tex]Let \ v = x; \ Let \ dv = dx[/tex]
    [tex]\int \frac{1}{2} \ln{(1 + x^2)} dx = \frac{1}{2} x \ln{(1 + x^2)} - \int \frac{x}{1 + x^2} dx[/tex]
    [tex]\int \frac{1}{2} \ln{(1 + x^2)} dx = \frac{1}{2} x \ln{(1 + x^2)} - \frac{1}{2} \ln{(1 + x^2)}[/tex]

    I then substitute the second integration into the first and obtain:

    [tex]\int x \arctan{x} dx = x (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) - \frac{1}{2} x \ln{(1 + x^2)} - \frac{1}{2} \ln{(1 + x^2)}[/tex]
    [tex]2 \int x \arctan{x} dx = x^2 \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}[/tex]
    [tex]\int x \arctan{x} dx = \frac{1}{2} x^2 \arctan{x} - \frac{1}{4} \ln{(1 + x^2)} + C[/tex]

    The answer in the back of the book is:

    [tex]\int x \arctan{x} dx = \frac{1}{2} (x^2 + 1) \arctan{x} - \frac{1}{2} x + C[/tex]

    I just can't seem to find out what I am doing wrong with my integration. I graphed both the book's answer and my answer on a graphing calculator and found my answer to be wrong by a little but not by much. Could somebody show me where my technique is flawed? I don't see how the book's answer doesn't contain any natural logarthmic functions. Thank you!
    -James
     
    Last edited: Aug 27, 2009
  2. jcsd
  3. Aug 27, 2009 #2
    Ok, obviously you messed up the things.

    First let u=arctan(x), and dv=x

    What are du and v?

    What do you think you should do next?

    Regards.
     
    Last edited: Aug 27, 2009
  4. Aug 27, 2009 #3
    Thanks for your help. Just after you responded I found my mistake. I have tried using the substitution you recommended before but I kept getting the wrong answer because of a small math error. Now I have found it. Thanks!
    -James
     
  5. Aug 27, 2009 #4

    Cyosis

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    Homework Helper

    The mistake is not so obvious in fact although cumbersome his first partial integration was correct. Also using your suggestion you would get [itex]\int \arctan x dx[/itex] yet the integral he needs to solve is [itex]\int x\arctan x dx[/itex]. The best way to do this would be u=arctan(x) and v=1/2 x^2.


    That said the OPs method works as well it is just more difficult. This is where you made the error:
    [tex]\int \frac{1}{2} \ln{(1 + x^2)} dx = \frac{1}{2} x \ln{(1 + x^2)} - \int \frac{x}{1 + x^2} dx[/tex]

    It should be:
    [tex]
    \frac{1}{2} x \ln{(1 + x^2)} - \int \frac{x^2}{1 + x^2} dx[/tex]
     
    Last edited: Aug 27, 2009
  6. Aug 27, 2009 #5
    Thank you. That helps a lot. I have been looking at that for a long time and just couldn't find the mistake.
     
  7. Aug 27, 2009 #6
    Yep, sorry I made typo. I meant u=arctan(x) and dv=x, so that v=x^2/2

    Regards.

    You're welcome! :smile:
     
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