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Indefinate Integral 1/(x^2 +1)

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data

    This is the very last piece of an integration using partial fractions question. I have to integrate 1/x2+1.

    2. Relevant equations

    Tan2x + 1 = Sec2x

    3. The attempt at a solution

    Substitute tanu=x:

    1/tan2u + 1 = 1/sec2u = cos2u

    which leaves me with int [ cos2u ] .dx I'm not sure what to do from here. Or if what I've done is correct.
     
  2. jcsd
  3. Jan 7, 2014 #2
    Do you know a function whose derivative is ##\frac{1}{x^2+1}##?
     
  4. Jan 7, 2014 #3

    D H

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    You haven't complete the u-substitution. You need to express dx as (something)*du. You used the u-substitution tan(u)=x, so what does that mean dx is?
     
  5. Jan 7, 2014 #4
    I'm not sure, that's the bit I'm having difficulty with.
     
  6. Jan 7, 2014 #5

    D H

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    What is ##\frac {d}{du} (\tan u)## ?
     
  7. Jan 7, 2014 #6
    I think I have it. I have

    d/dx(tanu)=d/dx(x)
    du/dx(sec2u)=1
    du/dx=1/sec2u
    dx=sec2u .du

    but I believe this to be incorrect, as that leaves me with

    ∫cos2x.sec2x.du = ∫1 .du = u.
     
  8. Jan 7, 2014 #7
    I believe that is sec2u, but I don't see how it relates to dx.
     
    Last edited: Jan 7, 2014
  9. Jan 7, 2014 #8

    LCKurtz

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    Aside from the missing constant of integration, that is correct. You started with ##x=\tan u##. What do you get if you solve for ##u## to get your answer in terms of ##x##?
     
  10. Jan 7, 2014 #9
    Ahh, I see now. arctanx=u. So ∫1/x2 + 1 .dx = arctanx +c. Thanks.
     
  11. Jan 7, 2014 #10

    Mark44

    Staff: Mentor

    You need to use parentheses when there are multiple terms in the top or bottom of a fraction.
    Write this as 1/(x2 + 1). What you wrote is (1/x2) + 1, which isn't what you meant.


    Both examples above should be 1/(tan2(u) + 1).
     
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