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Math Help - partial fraction decompositions

  1. #1
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    partial fraction decompositions

    So I have an assignment, and we are supposed to integrate using partial fraction decompositions. The integral is:

    (x^2 + 53)/(x^3 + x^2)

    What I have so far is:
    A/(x) + (Bx+C)/(x^2) + D/(x +1)

    Then:
    A(x^2 + x) + (Bx + C)(x+1) + D(x^2) = (x^2 + 53)

    I've determined that C= 53, and D= 54, but I'm stuck finding A & B. Any help would be much appreciated. Once I figure out what these partial fractions are, I'll be home free, since integrating is the easy part.

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jmaillou View Post
    So I have an assignment, and we are supposed to integrate using partial fraction decompositions. The integral is:

    (x^2 + 53)/(x^3 + x^2)

    What I have so far is:
    A/(x) + (Bx+C)/(x^2) + D/(x +1)

    Then:
    A(x^2 + x) + (Bx + C)(x+1) + D(x^2) = (x^2 + 53)

    I've determined that C= 53, and D= 54, but I'm stuck finding A & B. Any help would be much appreciated. Once I figure out what these partial fractions are, I'll be home free, since integrating is the easy part.

    Thanks
    one way: expand the left hand side:

    so we have: (A + B + D)x^2 + (A + B + C)x + C = x^2 + 53

    now equate the coefficients of like powers of x.

    from this we see:

    A + B + D = 1 ..............(1)

    A + B + C = 0 ...............(2)

    C = 53 .........................(3)

    since you know C and D, solving for A and B should be easy from this system ...oh wait, we still have a clue missing

    there might be some virtue in doing \frac {x^2}{x^2 (x + 1)} + \frac {53}{x^2(x + 1)}

    so the partial fractions are: \frac 1{x + 1} + \frac {53}{x^2(x + 1)}

    if you desire you can work on the second fraction a bit more to see if you can get partial fractions from it
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  3. #3
    Senior Member Peritus's Avatar
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    I think maybe the problem you had is that you have only 3 equations and 4 unknowns, and after finding D and C you get two identical equations:

    A+B = -53

    ------------------------------
    so you can choose the value of either A or B as you wish, for example a convinient choice would be B=0 and thus A=-53...
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Peritus View Post
    I think maybe the problem you had is that you have only 3 equations and 4 unknowns, and after finding D and C you get two identical equations:

    A+B = -53

    ------------------------------
    so you can choose the value of either A or B as you wish, for example a convinient choice would be B=0 and thus A=-53...
    yes, i saw that afterwards. i edited my post. don't know how much better i made the situation. too lazy to check it
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