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Basic Integration by parts

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data
    problem: [itex]\int[/itex]2arctanx dx
    2[itex]\int[/itex]arctan dx

    u=arctanx
    du=1/(1+x2)
    v=x
    dv=dx

    xarctanx-[itex]\int[/itex]x/(1+x2)

    integrate by parts a second time...

    u=x
    du=dx
    v=arctanx
    dv=1/1+x2

    xarctanx-[itex]\int[/itex]arctanx

    My final answer I get it 2xarctanx-2xarctanx+2/x2+1 which is obviously wrong. Thanks.
     
  2. jcsd
  3. Aug 28, 2011 #2

    rock.freak667

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    Homework Helper

    Don't go integration by parts here, just do a substitution of u=1+x2
     
  4. Aug 28, 2011 #3
    Thanks, I got the answer. This may seem like a dumb question, but how come integration by parts didn't work for this part?
     
  5. Aug 28, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    My final answer I get it 2xarctanx-2xarctanx+2/x2+1 which is obviously wrong. Thanks.[/QUOTE]
    Because [itex]2/(x^ 1) is the derivative of arctan(x), not the integral.

    Your choice of "u" and "dv" are just the results you got from the first integration by parts so you are just reversing the first integration. What you would correctly get is
    [tex]x arctan(x)- x arctan(x)+ \int arctan(x)dx= \int arctan(x)dx[/tex]
    exactly what you started with.
     
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