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Trigonometric / Mixed Integral Involving Substitution

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data

    integrate xarctanx/(x^2 + 1)^2

    2. Relevant equations

    Integration by parts possibly? I was attempting to do it without integration by parts because we went over this in tutorial and my TA couldn't solve it properly, hence why I'm here.

    3. The attempt at a solution

    I thought I could sub u = xarctanx so that du = 1/(x^2 + 1) dx but then I'm still left with one of the two x^2 + 1's. Wolfram gave me usinucosu du as the integral after substitution was complete and I feel like knowing how this was possible will be useful if I see any integrals like this in the future (rather than having to try integrating by parts).
     
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  3. Jul 27, 2011 #2

    SammyS

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    To find the derivative of x arctan(x), you need to use the product rule, so your result is incorrect. In fact, your result is the derivative of arctan(x) by itself.
    Did you understand any of what WolframAlpha did?
     
  4. Jul 27, 2011 #3
    I didn't understand much of what wolfram did, except in principle. I made a mistake though: It substituted u = arctanx alone, so du = 1/(x^2 + 1) which led it to the conclusion that this was equivalent to the integral usinucosu du. You're right that I will eventually have to use parts, and I think I can handle that if I just know how it got this substitution.
     
  5. Jul 27, 2011 #4
    But I personally don't see anywhere that usinucosu would have come from.
     
  6. Jul 27, 2011 #5
    Anyone?
     
  7. Jul 27, 2011 #6

    SammyS

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    [itex]\displaystyle \int\frac{x\arctan(x)}{(x^2+1)^2}\,dx[/itex]

    Let [itex]\displaystyle u=\arctan(x)\quad\to\quad du=\frac{1}{x^2+1}dx\,.[/itex]

    That takes care of everything except [itex]\displaystyle \frac{x}{x^2+1}\,.[/itex]

    Also, x = tan(u), so that gives us: [itex]\displaystyle \int\frac{\tan(u)\,(u)}{(\tan^2(u)+1)}\,du\,.[/itex]

    Use the trig identity tan2(u) + 1 = sec2(u) .

    Then simplify.
     
  8. Jul 27, 2011 #7
    Thank you so much, I really appreciate it! I need to make sure I realize that arctan(x) = u also says something about x (which should have been rather obvious) in the future.
     
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