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Homework Help: Integration by Partial Fractions Help

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    ∫ [x^(3)+4] / [x^(2)+4] dx

    2. Relevant equations

    3. The attempt at a solution
    I know that the fraction is improper, so I used long division to rewrite it as x+(-4x+4)/[x^(2)+4].
    Given the form S(x)+R(x)/Q(x), Q(x) is a distinct irreducible quadratic factor [x^(2)+4].
    I used the rule ax^2+bx+c ⇒ (Ax+B)/(ax^2+bx+c) to rewrite it as (Ax+B)/[x^(2)+4].
    I then solved for A and B and got A=-4 and B=4.
    I am now trying to solve ∫ [ x + (-4x+4)/(x^(2)+4) ] dx
    I know that ∫x=(1/2)x^2, but I am stuck with integrating (-4x+4)/[x^(2)+4].
    (I tried u-substitution and that didn't work.)
  2. jcsd
  3. Mar 18, 2015 #2


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    Hello StrangeCharm. Welcome to PF !

    Split ##\displaystyle \frac{-4x+4}{x^2+4}## into two fractions:

    ##\displaystyle \frac{-4x}{x^2+4} + \frac{4}{x^2+4}##

    The integral of one of them can be done via u-substitution. The other is a fairly well known integral.
  4. Mar 18, 2015 #3
    Thanks for the tip!
  5. Mar 19, 2015 #4
    I'm almost done with the problem but am having trouble with integrating 4/[x^(2)+4]. It looks like integrating 1/(1+x^2)dx and getting arctan(x); however, I'm not sure how to simplify the fraction into a similar form.
    Also, I'm wondering whether it is valid to make ∫4/(4+x^2)dx ⇒ ∫[2/(2+x)]^2dx and using u-substitution with u=x+2. I'm not sure because doing this would get a different result from the method above.
  6. Mar 19, 2015 #5


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    Is (2 + x)2 = 4 + x2 ? No.

    Use u substitution: u = 2x .
  7. Mar 19, 2015 #6
    Okay, right... I just realized that. Thanks for the help!
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