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Integration by long division problem

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the indefinite integral of (4 x^3+4 x^2-96 x -100)/(x^2-25)dx

    Then the integrand decomposes into the form

    ax + b + c/(x - 5) + d/(x + 5)

    Find a, b, c, and d.

    Then find the integral of the function.

    3. The attempt at a solution

    Using long division, I got this far...

    [tex]\frac{4x^3 + 4x^2 - 96x - 100}{x^2 - 25}[/tex]

    =

    4x + 4 + c/(x - 5) + d/(x + 5)

    It'd be pretty hard to show how I got a and b, but I'm pretty positive that's correct. I just can't find C or D.

    So then, I thought I was supposed to put my remainder after dividing over (x - 5)(x + 5). My remainder is 4x, so I went.

    4x/((x - 5)(x + 5))

    Which produces

    c + d = 4.

    I need one more equation to solve for it though. Help?
     
  2. jcsd
  3. Mar 16, 2008 #2

    tiny-tim

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    … just multiply …

    Hi the7joker7! :smile:

    c/(x - 5) + d/(x + 5) = 4x/(x - 5)(x + 5);

    just multiply both sides by (x - 5)(x + 5). :smile:
     
  4. Mar 16, 2008 #3
    So c + d = 4 is correct?

    So that gets me to...

    C(x + 5) + D(x - 5) = 4x

    Cx + 5c + dx + 5d = 4x

    4x = (c + d)x + (5c + 5d)

    Again, produces c + d = 4

    So that doesn't really help. =/
     
  5. Mar 16, 2008 #4

    cristo

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    Basically, you want to solve this equation:
    [tex] \frac{4x^3+4x^2-96x-100}{x^2-25}=ax+b+\frac{c}{x-5}+\frac{d}{x+5}[/tex]

    Now, the way I would do it would be to multiply up the right hand side, and put it all over the common denominator (x-5)(x+5). Then, you can compare the numerator of the LHS to the new numerator on the RHS. Compare coefficients: you correctly have that a=b=4. Comparing the coefficients of the x and units will give you two equations, in terms of c and d, which you can solve.
     
  6. Mar 16, 2008 #5

    tiny-tim

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    … gotta be careful …

    erm … no … Cx + 5c + dx - 5d = 4x! :redface:

    Try again! :smile:
     
  7. Mar 16, 2008 #6
    C + D = 4

    and

    5c - 5d = 0, then?

    So that gives...

    c = 2 and d = 2?

    I just had another guy claim it was c = 4 and d = 0 using...

    "using synthetic division divide numerator by denominator

    x^2-25)4x^3+4x^2-96x-100(4x
    ...........4x^3+0x^2-100x
    ______________________
    ...................4x^2+4x-100(4
    ...................4x^2+0x-100
    _______________________
    ...........................4x

    so (4x^3+4x^2-96x-100) /(x^2 - 25) = 4x + 4 + 4x/(x^2-25)

    so a = 4, b = 4 , c = 4 and d = 0"

    Which one's right?
     
  8. Mar 16, 2008 #7

    tiny-tim

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    Hi joker! :smile:

    Well, he's sort-of right, and he sort-of isn't!

    The first line above is correct - but it's exactly what you had anyway!

    It isn't reduced to the simplest fractions!

    So the second line is just optimistically re-defining c and d to fit the result!

    You've gone one step further, and split the last fraction into two simpler ones.

    Yours is defintitely right1 :smile:
     
  9. Mar 16, 2008 #8
    Thanks.

    Hmm...Since derivative of denominator (x^2-25) is numerator, 2x , the integral of 2x dx/(x^2-25) is ln(x^2-25)

    =>4x^2/2 + 4x + 2ln(x^2 -25) + c

    =>2x^2 + 4x + 2ln[(x+5)(x-5)] + c

    2x^2 + 4x + 2ln(x+5) + 2ln(x-5) + c

    Does that sound in order?
     
  10. Mar 16, 2008 #9

    tiny-tim

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    Hi joker! :smile:

    Yes that's fine! :smile:

    btw, I think the point "another guy" was correctly making was that, once you'd got 4x/(x^2 - 25), there was no point in breaking it down any further, since you could instantly see what its integral was!

    But you had to go on in this case only because the question specifically required it.
     
  11. Jun 9, 2009 #10
    If it helps any...

    Once you get your result from long division (should be 4x+4-(4x/x^2-25)), multiply through by x^2-25 on both sides. This will leave you with 4x^3+4x^2-96x-100 = 4x(x^2-25)+4(x^2-25)-4x.

    Now set x = 0 , so A = 4

    Now set x = 5 , so C = -1

    Now set x = 1 , so B = -2 ( don't forget to plug in A and C to get B)

    So your A B and C should be 4,-1, and -2 respectively.
     
  12. Jun 9, 2009 #11
    You don't need to solve a single equation.

    f(x) = (4 x^3+4 x^2-96 x -100)/(x^2-25)

    Expand around singular points and find asymptotic behavior at infinity:

    Singular term in expansion around x = -5:

    1/(x+5) * Lim x--->-5 of (x+5)f(x) = 2/(x+5)

    Singular term in expansion around x = 5:

    1/(x-5) Lim x--->5 of (x-5)f(x) = 2/(x-5)

    Expansion around infinity:

    The singular terms in this expanson (i.e. the terms that go to infinity as we approach the point around which we expand) are the postive powers of x. We have:

    1/(x^2 - 25) = 1/x^2 1/[1-(5/x)^2] = 1/x^2 [1+25/x^2 + ...]

    f(x) = 4 x + 4 + terms that go to zero for x to infinity

    The sum of all the singular terms of the three expansions is:

    g(x) = 2/(x+5) + 2/(x-5) + 4 x + 4

    Consider the difference f(x) - g(x). Since both f and g are rational functions, f(x) - g(x) is a rational function. But it doesn't have any singularities as g contains the singular terms of the expansion around all the singular points of f. So, f(x) - g(x) is actually a polynomial. At infinity, f(x) - g(x) must tend to zero, as g(x) contains the positive powers of x in the large x expansion of f(x). This then implies that
    f(x) - g(x) is zero.
     
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