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Calculus Integration Problem

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Int x/x^4+x^2+1 dx

    2. Relevant equations

    u=x^2 dx=2x

    3. The attempt at a solution
    I tried u=x^2+x+1 then du=2x+1
    1/2 int u+1/u^2 du
    I thought I had it because it made a nice u/u^2 and 1/u^2
    But no dice says the prof.
    He said let u=x^2 that gives me 1/2 int du/u^2+u+1/u but I'm stuck.
    Maybe partial fractions?
     
    Last edited: Oct 28, 2009
  2. jcsd
  3. Oct 28, 2009 #2

    lanedance

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    hey imataxslave, welcome to pf - you should try & use brackets to make it clear what your equations actually are & make it easy for people to read

    ie x/x^4+x^2+1 = (1/x^3 + x^2 +1) to me, so you should write it as
    x/(x^4+x^2+1)

    though guessing at what you have done, you get to
    [tex] \int \frac{du}{u^2 + u + 1} [/tex]

    now i haven't tried it, but you could try completing the square on the denominator, then substututing for whatever is in the the square part - should hopefully take it to a more familiar form
     
  4. Oct 28, 2009 #3

    lanedance

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    updated above
     
  5. Oct 28, 2009 #4
    Thanks. Sorry about the mess. This is only my fourth math class. Now that I'm sharing with others I've been getting many requests to quit being so sloppy. I tried to complete the square but the one cancels out and I' back to the original input. I think I remember a trick using fractions, I'll look into it. This at least gives me a direction to go.
     
  6. Oct 28, 2009 #5
    thanks for the idea! Problem solved.
    let u =x^4+x^2+1
    du= 4x^3+2x dx
    Complete the square
    (x^2)^2+(2x^2/2)+1/4+1-1/4 dx
    =int x/(((x^2+1/2)^2)+(3/4)) dx
    let u^2=(x^2+1/2)
    du=2xdx
    =int 1/2 du/(u^2+3/4)
    =1/2 int du/(u^2+(sqrt3/2)^2)
    that's the form I need because dx/(x^2+a^2)= (1/a)arctan (x/a) + c
    all that's left is plugging it in and changing back substitutions.
    I hope this helps others. It's going in my book of half day problems.
     
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