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Improper Integral

  1. Apr 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral: [tex]\int[/tex][tex]\frac{dx}{x^{3}+x^{2}+x+1}[/tex]
    from infinity to zero

    2. Relevant equations
    lim t--> infinity [/tex] [tex]\int[/tex] [tex]\frac{dx}{x^{3}+x^{2}+x+1}[/tex]


    3. The attempt at a solution

    lim t-->infinity [/tex] [tex]\int[/tex] [tex]\frac{dx}{(x+1)(x^{2}+1}[/tex]


    I'm stuck on where to go from here. I tried partial fractions, but can't seem to get it. any hints would be a great help!
     
  2. jcsd
  3. Apr 3, 2008 #2
    Hi,
    partial fractions, yes.

    [tex]
    \frac{1}{x^3+x^2+x+1}=\frac{A}{x+1}+\frac{B+Cx}{x^2+1}=\frac{A(x^2+1)+(B+Cx)(x+1)}{x^3+x^2+x+1}
    [/tex]

    So you must have
    [tex]
    1=A(x^2+1)+(B+Cx)(x+1)=(A+C)x^2+(B+C)x+A+B
    [/tex]

    Comparing coefficients of the same powers of x you get the equation:


    1=A+B
    0=B+C
    0=A+C

    which you can easily solve, I assume :smile:

    Do you know to integrate the partial fractions?
     
    Last edited: Apr 3, 2008
  4. Apr 3, 2008 #3
    Oh! I see, i must have miswritten something when i was doing partial fractions. Thank you so much for the help!

    I 'll give it a shot and see what comes up
     
  5. Apr 3, 2008 #4
    alright, so I've worked on solving this problem up to:

    a=1/2 b=1/2 c=-1/2

    so my integral terms would be:
    [tex]\frac{1/2}{x+1}[/tex]-([tex]\frac{(1/2)x-(1/2)}{x^{2}+1}[/tex])

    taking the antiderivative:
    i have, [tex]\frac{1}{2}[/tex]ln|x+1| for the first term
    as for the second, i know one of the terms will be tan[tex]^{-1}[/tex]x because of the denominator, but i'm having troubles with the numerator since I can't use substitution for it.
     
  6. Apr 3, 2008 #5
    Split the second term into two. For the one with the x in the numerator you can use the substitution


    u=x^2
    du=2xdx

    The first term (with the constant numerator).. well..you know how to do it:smile:
     
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