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Integration - Can't find my mistake

  1. Jan 14, 2014 #1
    ##\int arctan(\frac {1}{x})dx = I## when ##u = arctan(\frac {1}{x}), du = -\frac {1}{1+x^2} dx, dv = dx, v = x## and I got ##du## by chain rule since ##\frac {1}{1 + \frac {1}{x^2}} (-\frac {1}{x^2})## simplifies to ##-\frac {1}{1+x^2} dx## right?

    ##I = xarctan(\frac {1}{x}) + \int \frac {x}{1+x^2}dx##
    ##I = xarctan(\frac {1}{x}) + \int \frac {1}{x}dx + \int xdx##
    ##I = xarctan(\frac {1}{x}) + ln lxl + 0.5x^2 + C ##

    This is not the correct answer, though, and I can't seem to find my exact error.

    Any help would be great!

    P.S. I have seen the identities to make arctan(1/x) = arctan(x) and using substation earlier, but am trying to use this method.

    Second integration problem:

    ##I = \int \frac {r^3}{4+r^2}^0.5 dr## and if ##4 +r^2 = u## or ##r^2 = u - 4## then ##dr = \frac {1}{2r} du## so
    ##I = 0.5 \int \frac {u-4}{u}^0.5 du = 0.5 [\int u^0.5 - 4\int {1}{u}]##
    ##I = \frac {1}{3} u^{3/2}- 8u^{1/2}##

    This is also wrong but I'm not quite sure why.
     
    Last edited: Jan 14, 2014
  2. jcsd
  3. Jan 14, 2014 #2

    LCKurtz

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    ##\frac x {1+x^2} \ne x + \frac 1 x##.
     
  4. Jan 14, 2014 #3
    General case: ##\frac {d}{dy} arctan(y) = \frac {1}{1 +y^2}## right?
    So replacing y with 1/x
    ##= \frac {1}{1 + (1/x)^2}##
    ##= \frac {1}{\frac {x^2 + 1}{x^2}}##
    ##= \frac {x^2}{1+x^2}## [1]

    Then applying chain rule to ##\frac {1}{x}##
    ##\frac{d}{dx}\frac {1}{x} = -\frac {1}{x^2}## [2]

    So I multiply [1] by [2] to get

    ## -\frac {1}{1 + x^2}##

    Oh, never mind, just realized you stated something completely different...
     
  5. Jan 14, 2014 #4
    Just a question: If this was a definite integral and did not included 0 between the upper and lower limits, would it be allowed or still no? Since you have to find the general antiderivative either way, I guess it doesn't make sense...

    Just wondering, what exactly would be the next step to take here? I seem to have made the same mistake in both questions.
     
  6. Jan 14, 2014 #5

    LCKurtz

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    You had$$
    x\arctan(\frac 1 x) +\int\frac x {x^2+1}~dx$$Do a ##u## substitution on the second integral.
     
  7. Jan 14, 2014 #6

    SteamKing

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    Your integral expression appears corrupted.

    Are you perhaps trying to find

    I = [itex]\int (\frac {r^3}{4+r^2})^{0.5} dr[/itex]
     
  8. Jan 14, 2014 #7
    Sorry, yes. I was in a hurry and forgot to fully check my script. My problem has been solved by the above answer. Thank you for all the help!
     
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