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Integral : (x^2)/((x^2+1)^2)

  1. Dec 25, 2009 #1
    The answer of the integral of (x^2)/((x^2+1)^2) is (1/2)(arctan(x)-(x/x^2+1))

    In class, we've seen the steps to solve this integral, but I don't understand certain steps..
    Can someone explain me how to solve this integral, step by step?
  2. jcsd
  3. Dec 25, 2009 #2


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    If you can post the steps and point out those that you didn't understand then I'm sure someone can help you.

    BTW. The easiest way to do that one is "integration by parts". Have you learnt this technique yet?
  4. Dec 25, 2009 #3
    Sure, no problem, here are the steps:

    = (1/2)*Integral(x d(1/(x^2+1))
    = (1/2)*(x/(x^2+1))-(1/2)*Integral(1/(x^2+1)dx)
    = (1/2)*(x/(x^2+1))-(arctan(x))/2

    Last step is the answer

    (The answer I said earlier was wrong, this is the correct answer:(1/2)*(x/(x^2+1))-(arctan(x))/2)

    Thank you
  5. Dec 25, 2009 #4
    [tex]\int\frac{x^2}{(x^2 + 1)^2} = \int x \frac{x}{(x^2 + 1)^2}[/tex]

    Using ∫u v' = uv - ∫v u',
    let u = x and v' = x/(x2 + 1)2
  6. Dec 25, 2009 #5
    It's basically separating it into parts ie.

    [itex]\int \frac{x^2}{(x^2+1)^2}\rightarrow \int \frac{x}{1}.\frac{x}{(x^2+1)^2}\equiv x(x. \sin(\arctan(x)))[/itex]



    and [itex]x\frac{x}{(1+x)^2}=x.\sin(\arctan(x))[/itex]

    By the trig identity.

    Thus the answer is:


    Don't forget the constant of integration, it's a silly way to loose marks. :smile:
    Last edited: Dec 25, 2009
  7. Dec 25, 2009 #6
    Thank you both!
  8. Dec 25, 2009 #7
    np Bhorok's answer is more elegant and easier, but I thought you might need a long winded explanation and there's often more than one way to swing a cat I guess. Hope it helped. :smile:
  9. Dec 25, 2009 #8
    since you have the answer, take its derivative & work backwards. that's how to figure it out. just don't show anyone your rough work :tongue2:
  10. Dec 25, 2009 #9


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    No the original answer was correct, you dropped a minus sign in the first line of this derivation.
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