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Integration by partial fractions

  1. Mar 2, 2008 #1
    [SOLVED] integration by partial fractions

    1. The problem statement, all variables and given/known data


    2. Relevant equations

    A1/ax+b + A2/(ax+b)^2 + .... + An/(ax+b)^n

    3. The attempt at a solution

    (2x^2-1)/(4x-1)(x^2+1) = A/4x-1 + Bx+C/x^2+1

    2x^2-1/x^2+1 = A + Bx+C(4x-1)/x^2+1

    set x = 1/4, get

    -14/17 = A + B(0), A = -14/17

    I'm pretty much stuck here; I just need to get the values for B and C and I can solve it from there, any help is greatly appreciated.
  2. jcsd
  3. Mar 2, 2008 #2
    I've found the value for C:

    2x^2-1/x^2+1 = -14/17 + 4Bx^2-Bx+4Cx-C/x^2+1

    set x = 0, get

    -1 = -14/17 -C, -3/17 = -C, C = 3/17
  4. Mar 2, 2008 #3
    I find it more usefull to put the right hand side over a common denominator
    [tex]{A(x^{2}+1) + (Bx+C)(4x-1)}\over{(4x-1)(x^{2}+1)}[/tex].

    Now expand the numerator on both sides of the equation (i.e. your first equation in part 3.) and group the like powers of x.
    Then the coefficient of each power must match on either side of the equal sign. Matching the coefficients should give you a system of 3 equation that you can solve for A, B, and C.
  5. Mar 2, 2008 #4
    yeah I know there are 2 methods to do integration by partial fractions, the standard method which you're describing and another method called heathside which I've been using. I'm more comfortable with the standard method but the heathside method seems to be working better for the problems my professor assigned.

    I tried the grouping like powers of x, but it left me with this:

    2x^2-1 = Ax^2 + A + 4Bx^2-Bx+4Cx - C

    it left me with the -Bx + 4Cx with no like power to go along with.
  6. Mar 2, 2008 #5
    When the corresponding power is missing, that means the coefficient is zero.
    The coefficient of x is zero, so 4C-B=0 would be your missing equation.
  7. Mar 2, 2008 #6
    wow! I never realized that; the damn book doesn't explain specific cases like that; thanks for the help, I'll post the result here in a minute.
  8. Mar 2, 2008 #7
    wow, I hate this section:

    [tex]\int[/tex]((-14/17)/(4x-1))dx + [tex]\int[/tex]((12/17x + 3/17)/(x^2+1))dx

    for the 1st integral:

    u = 4x - 1

    du = 4dx or 1/4du = dx

    so (-14/17)/4[tex]\int[/tex](du/u) = -7/34ln(4x-1)

    for the 1st half of the 2nd integral


    u = x^2+1, du = 2xdx or 1/2du = xdx

    --> (12/17)/2[tex]\int[/tex]du/u = 6/17ln(x^2+1)

    for the last half of the 2nd integral

    3/17[tex]\int[/tex](dx/x^2+1) = arctanx + C

    so the whole thing equals out to:

    -7/34ln(4x-1) + 6/17ln(x^2+1) + 3/17arctanx + C

    thanks for the help man.
  9. Mar 2, 2008 #8


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    Personally, I'd simplify it first by saying:
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