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Homework Help: Integration of Rational Function - problem

  1. Nov 25, 2008 #1
    1. The problem statement, all variables and given/known data

    Here's the question: int(1/(x(x^2+3)*sqrt(1-x^2)))dx

    2. Relevant equations

    According to the textbook, the answer should be: (1/3)*ln((1-sqrt(1-x^2))/x)+(1/12)*ln((2+sqrt(1-x^2))/(x^2+3))+C

    3. The attempt at a solution

    1) let t = sqrt(1-x^2), so dt = (-x)/sqrt(1-x^2) dx
    2) substituted t and dt into the equation and got the following: -int(1/((1-t^2)(4-t^2)))dt
    3) expanded the denominator into (1+t)(1-t)(2+t)(2-t) and used partial fractions to find A,B,C and D in the numerator (respectively)

    Using partial fractions, I got: A & B = 1/6, C & D = -1/12

    4) From there, I integrated all four portions separately to yield:
    -(1/6)*ln(1+t)-(1/6)*ln(1-t)+(1/12)*ln(2+t)+(1/12)*ln(2-t)+C

    5) Substituting t = sqrt(1-x^2) back into the equation and collecting like terms, I got:
    -(1/6)*ln((1+sqrt(1-x^2))(1-sqrt(1-x^2)))+(1/12)*ln((2+sqrt(1-x^2))(2-sqrt(1-x^2)))+C

    6) Finally, multiplied everything inside both ln's and brought out the 2 in the first term (ln(x^2)) to get 1/3:
    My final answer: -(1/3)*ln(x)+(1/12)*ln(x^2+3)+c

    Thank you for your help in advance!!!!!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 25, 2008 #2

    Mark44

    Staff: Mentor

    Your work shows me that you know what you're doing, but I believe you made an error in applying your substitution. You have t = sqrt(1 - x^2), from which we get t^2 = 1 - x^2, so x^2 = 1 - t^2, and finally x = sqrt(1 - t^2).

    When I did the substitution, I got this:
    [tex]\frac{-dt}{\sqrt{1 - t^2} (4 - t^2) t}[/tex]

    IOW, I didn't get a factor of (1 - t^2) as you did; I got a factor of sqrt(1 - t^2), and a factor of t. Those make for a different problem.
     
  4. Nov 26, 2008 #3

    HallsofIvy

    User Avatar
    Science Advisor

    That is not, by the way, a rational function.
     
  5. Nov 26, 2008 #4
    Thanks for your help Mark44! I was on the right track but had a slight integration error that mixed up my +/- signs. Plus, I found out that the textbook answer was expressed with the absence of sqrt's in the denominator - conjugates!!
     
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