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Rational Function Integration

  1. Apr 18, 2008 #1
    I want to integrate:

    1/[(x + 1)*(x^2 + x +1)] dx

    Now the quadratic has complex routes, and we have not done any integration with that yet, so I broke it up into its partial fractions.

    A/(x +1) + (Bx + C)/(x^2 + x +1)

    But I cannot seem to find the numbers A B C. mamybe I am just missing something real obvious?? Any pointers in the right direction? Cheers guys.

    PS. Is the proof of the theory that you can break up fractions like that beyond a first year math for science course?
     
  2. jcsd
  3. Apr 18, 2008 #2

    HallsofIvy

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    I have no idea what a "math for science" course is, but the proof that there exist numbers A, B, C that will work doesn't involve anything more than basic algebra.

    Here you want to find A, B, C so that
    [tex]\frac{1}{(x+1)(x^2+ x+1)}= \frac{A}{x+1}+ \frac{Bx+ C}{x^2+ x+ 1}[/tex]
    Multiply both sides by (x+1)(x2+ x+ 1) to get
    [tex]1= A(x^2+ x+ 1)+ (Bx+ C)(x+1)[/tex]
    If you let x= -1, that reduces to 1= A.

    I expect you had already done that. The problem is that that there is no value of x that makes x2+ x+ 1= 0. You cannot "reduce" the equation that easily but since the equation is true for all x, you can still get two equations for B and C by letting x be any number you want. I would suggest putting x= 0 and x= 1 into the equation.
    If x= 0, the equation becomes 1= A+ C and you already know A. If x= 1, the equation becomes 1= 3A+ (B+ C)(2) and you already know A and C.

    Another method that always works is to multiply out the right side and combine "like powers"
    [tex]1= A(x^2+ x+ 1)+ (Bx+ C)(x+1)[/tex]
    [tex]1= Ax^2+ Ax+ A+ Bx^2+ Bx+ Cx+ C[/tex]
    [tex]1= (A+ B)x^2+ (A+ B+ C)x+ (A+ C)[/tex]
    Since that must be true for all x, corresponding coefficients must be the same: A+ B= 0, A+ B+ C= 0, A+ C= 1.

    To integrate the term with x2+ x+ 1 in the denominator, complete the square to get (x+ 1/2)2+ 3/4 and let u= x+ 1/2.
     
    Last edited: Apr 18, 2008
  4. Apr 18, 2008 #3
    Thank you.

    I had already got A=1, and C=0. I don't know how I didn't get B. I guess I was a bit tired and lost track :shy:

    I was just wondering about the other part. My Mathematics course is part of a science course, so sometimes proofs aren't done, like they would be in a pure math course.
     
  5. Apr 18, 2008 #4
    On this question again, how do I integrate the second part if there is also an x in the top?
     
  6. Apr 18, 2008 #5

    HallsofIvy

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    As I said before, complete the square in the denominator so it is (x+ 1/2)2+ 3/4, then let u= x+ 1/2. You will have something of the form [itex](u+ c)/(u^2+ 3/4)[/itex]. Separate that as [itex]u/(u^2+ 3/4)+ c/(u^2+ 3/4)[/itex]. The first is easy: let v= u2+ 3/4. For the second remember that the derivative of arctan(x) is 1/(x2+ 1).
     
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