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[Integration] A tough substitution problem

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data

    The following is to be evaluated using substitution or partial integration.

    [tex] \int\frac{(2x-1)}{e^{\arctan(x)}}\,dx[/tex]

    (It's supposed to be e^(arctan(x)) but i'm new to LaTeX and can't quite figure out how I would input it correctly) (Fixed it for you.)

    2. Relevant equations

    No relevant equations

    3. The attempt at a solution

    I tried partial integration and that definitely does not help. I simply have no idea how I could substitute.
     
    Last edited by a moderator: Feb 1, 2012
  2. jcsd
  3. Feb 1, 2012 #2

    lanedance

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    how about trying integration by parts first, and let u' = 2x-1
     
  4. Feb 1, 2012 #3

    lanedance

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    actually i might have to rethink that...
     
  5. Feb 1, 2012 #4
    Imho it makes sense to substitute u=arctan(x). From there, there is still some way to go though.
     
  6. Feb 1, 2012 #5

    vela

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    You just have to put everything in the exponent inside curly braces to group them together. Also, use a backslash before the name of common functions to get them to typeset correctly.

    There's a LaTeX guide here: https://www.physicsforums.com/showthread.php?t=546968
     
  7. Feb 1, 2012 #6

    lanedance

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    or noting that [itex] \frac{d}{dx}arctan(x)= \frac{1}{1+x^2}[/itex] you could rearrange as follows

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

    which should avoid subtitution or parts all together.. though it comes from a similar idea to parts
     
  8. Feb 2, 2012 #7
    where did you carry out the integral?
     
  9. Feb 2, 2012 #8

    lanedance

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    haven't done the integral, as i left that part for the OP, didn't mean to have the integral sign at the start of that expression, was rearranging to simplify the integration
     
    Last edited: Feb 2, 2012
  10. Feb 2, 2012 #9

    lanedance

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    So the integral becomes

    [tex]
    \int \left( 2xe^{-arctan(x)}-(1+x^2)\frac{e^{-arctan(x)} }{1+x^2} \right)dx
    [/tex]
     
  11. Feb 2, 2012 #10
    It can be verified that (1+x^2)/e^atan(x) is the answer, hope that helps
     
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