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How to integrate int 1/(4+x^2)^2 dx

  1. Jul 29, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    I understand that [tex]\frac{dx}{a^2+u^2}[/tex] = [tex]\frac{1}{a}[/tex] tan -1 [tex]\frac{u}{a}[/tex] + c

    3. The attempt at a solution
    The extra square is throwing me off for some reason. If I let a=2 and u=x it doesn't seem to help because of the whole term being squared. I end up with [tex]\frac{1}{2}[/tex]tan-1[tex]\frac{x}{2}[/tex] + c. In my book the first fraction is [tex]\frac{1}{16}[/tex]I'm pretty sure I can't complete the square because it isn't ax^2+bx+c. Basically I am stumped.
  2. jcsd
  3. Jul 29, 2009 #2


    Staff: Mentor

    Re: Integrals

    I would go with a trig substitution, tan u = x/2.
  4. Jul 29, 2009 #3
    Re: Integrals

    Yeah I can't believe I missed that substitution. I was going through my notes and realized I let that square block my train of thought for some reason. Its amazing what a small break and clearing your head can do.
    I just let a=2 and x=2 tan theta. What I forgot to do was take the derivative of x. Plugging it all back together and then using some right angle trigonometry I found out the problem was much easier than I though.
    Thanks for the help.
  5. Jul 29, 2009 #4
    Re: Integrals

    This is a good substitution technique to internalize. Often when you have a2+x2 in the denominator, you will make the substitution x=a*tan(u). If you have a2-x2, then you make either x=a*cos(u) or x=a*sin(u).
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