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Integration of a quotient with a factor in the denominator that has no real root

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data

    $$\int \frac{x-1}{(x+1)(x^2+1)} dx$$

    2. Relevant equations

    N/A

    3. The attempt at a solution
    I thought that I would use partial fractions, so:

    $$\frac{x-1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{B}{x^2+1}$$
    $$x-1 = A(x^2+1) + B(x+1)$$
    ##x=-1 \Rightarrow (-1)-1 = A((-1)^2+1) + B((-1)+1) = 2A \Rightarrow A = -1##

    I don't think my instructor wants me to use complex numbers, but ##\forall x\in\mathbb R, x^2+1 \neq0## and ##x^2+1## is irreducible in ##\mathbb R##.
     
    Last edited: Oct 22, 2012
  2. jcsd
  3. Oct 22, 2012 #2

    Mark44

    Staff: Mentor

    Your decomposition is incorrect. For the second fraction on the right side, use
    $$\frac{Bx + C}{x^2 + 1} $$
     
  4. Oct 22, 2012 #3
    Okay, if I use that, then
    $$\frac {x-1}{(x+1)(x^2+1} = \frac {A}{x+1} + \frac {Bx+C}{x^2+1} \Rightarrow x-1 = A(x^2+1) + (Bx+C)(x+1),$$
    but won't I still need to use the imaginary number ##i##?
     
    Last edited: Oct 22, 2012
  5. Oct 22, 2012 #4

    SammyS

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    Staff Emeritus
    Science Advisor
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    Gold Member

    You shouldn't need to use the imaginary unit, i .

    Simply equate the constant terms and equate coefficients of x and equate coefficients of x2
     
  6. Oct 22, 2012 #5
    If I do you use ##i##, then
    $$x=i \Rightarrow (i)-1 = A((i)^2+1) + (B(i)+C)((i)+1)$$
    $$ = A((-1)+1) + (Bi^2+Bi+Ci+C) = (0) + (-B+Bi+Ci+C)= (B+C)i + (C-B).$$
    Two complex numbers are equal only if both their real and imaginary parts are equal, so
    $$Im(i-1) = 1 = B+C = Im((B+C)i + (C-B)), Re(i-1) = -1 = (C-B) = Re((B+C)i + (C-B)).$$
    The system of equations
    $$1 = B+C$$
    $$-1= -B+C$$
    has solutions ##C=0, B=1##. Then,
    $$\int \frac{x-1}{(x+1)(x^2+1)} dx = -\int \frac{dx}{x+1} + \int \frac{x}{x^2+1} dx = -\ln|x+1|+\frac{1}{2}\ln|x^2+1|+C = ln\frac{\sqrt{x^2+1}}{x+1}+C$$

    Does this look right?

    I don't quite understand.

    Actually, I think I understand now.

    $$x-1=A(x^2+1)+(Bx+C)(x+1)=Ax^2+A+(Bx^2+Bx+Cx+C)=(A+B)x^2+(B+C)x+(A+C)$$
    Noticing that ##x-1=0x^2+x-1##, I get the following equalities
    $$0=A+B$$
    $$1=B+C$$
    $$-1=A+C.$$
    If ##A=-1##, then I get the same values for B and C and the same answer to this question as I did when I used ##i## to solve this question.
     
    Last edited by a moderator: Oct 22, 2012
  7. Oct 22, 2012 #6

    Mark44

    Staff: Mentor

    Why don't you check it yourself? If you differentiate what you ended up with, you should get your integrand.
    When you solve for the constants (A, B, and C in this case), the equation x−1=A(x2 +1)+(Bx+C)(x+1) is identically true for all values of x. The only way two polynomials, x - 1, and the one on the right, can be equal is for their corresponding coefficients to be equal. Since x - 1 doesn't have an x2 term (meaning that the coefficient of x2 is 0), the coefficient of x2 on the right has to be 0 as well. On the left, the coefficient of x is 1, so the coefficient of x on the right also has to be 1. Finally, the constant term on the left is -1, which must be the same as the constant term on the right.

    You have 3 variables, A, B, and C. Use any values of x to get three equations in these unknowns. Using i is probably more work than it's worth.
     
    Last edited: Oct 22, 2012
  8. Oct 22, 2012 #7
    Yeah, I realized that, but it took a while to edit my previous post. Thank you for your help. :)

    And it all works out when I differentiate my answer! Thanks again.
     
    Last edited: Oct 22, 2012
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