1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rational integral

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data
    the indefinite integral of x/(x^4+x^2+1)

    2. Relevant equations

    3. The attempt at a solution
    I didn't see an obvious u-substitution and it didn't look like a partial fractions candidate to me since the bottom is not easily factored. It doesn't look like any of the inverse trig forms. I didn't see how it could be integrated by parts, so I was pretty much lost.

    I tried letting u = x^2 so that du would = 2xdx in order to get rid of the x in the numerator, but that didn't really help me as I was left with 1/(u^2+u+1) and that isn't really much better than the original problem

    I also tried completing the square in the denominator to get (x^2+1)^2-x^2 my idea was to then allow u to =(x^2+1) and du would equal 2xdx. this left me with the exact same integral as before (of course) and didn't help.

    I'm stuck. I'm almost sure that there has to be a substitution I can use to get the integral, but I haven't the faintest idea what it is. I can't see anything that looks promising.
  2. jcsd
  3. May 13, 2007 #2
    Really? Why not try completing the square at this stage?
  4. May 13, 2007 #3
    because if I completed the square at this stage, I'd get the indefinite integral of du/(u+1)^2-u, right? If I do that, I don't see this would help as it still isn't an inverse trig form and I still don't see any candidates for another substitution...
  5. May 13, 2007 #4
    [tex]u^2+u+1[/tex] can also be written as [tex]\left(u+\frac{1}{2}\right)^2 + \frac{3}{4}[/tex]

    In such cases, the "trick" in completing the square, given a quadratic polynomial, is to divide the coefficient of the middle term (when it is written in the usual form ax^2+bx+c) by 2 and then subtracting (b/2)^2 from the constant c.
    Last edited: May 13, 2007
  6. May 13, 2007 #5
    ahh, wow, that works MUCH better. so that basically leaves me with an inverse tangent, right? good deal. Thanks
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook