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Math Help - Integration problem (partial fractions)

  1. #1
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    Integration problem (partial fractions)

    Integrate:

    \int \frac{2x^5 - 162x + 3}{x^4 - 81}dx

    Now I used partial fractions:

    \frac{2x^5 - 162x + 3}{x^4 - 81} = \frac{A}{x - 3} + \frac{B}{x + 3} + \frac{Dx + E}{x^2 + 9}

    and after a lot of working out got the following equations:

    A + B + D = 0

    3A - 3B + E = 12

    9A + 9B - 9D = -162

    <br />
27A - 27B - 9E = 3

    Now I tried gauss reducing it, but it was rather long and tedious which gave me:

    A = \frac{-167}{36}

    B = \frac{157}{-36}

    D = 9

    E = -\frac{1}{6}

    In my final answer I got:

    \frac{-167}{36}ln|x - 3| - \frac{157}{36}ln|x + 3| + \frac{9}{2}ln|x^2 + 9| - \frac{1}{54}arctan(\frac{x}{3}) + C

    which was of that right form (due to the logs and arctan) but obviously the values were wrong.

    So I was wondering, is there a more efficient way to do this question and solve for A, B, D and E without using gauss reduction?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Well, I don't know if you can use that, but I use some sort of inspection to get the values of A and B.

        \dfrac{2x^5 - 162x + 3}{x^4 - 81} = \dfrac{A}{x - 3} + \dfrac{B}{x + 3} + \dfrac{Dx + E}{x^2 + 9}

    Take the fraction:

        \dfrac{2x^5 - 162x + 3}{(x+3)(x-3)(x^2 + 9)}

    Now, what value do you have to assign to x so that one denominator becomes 0?

    We put x = 3.

    So, you find this:

     \dfrac{2x^5 - 162x + 3}{(x+3)(x^2 + 9)}

    And put x = 3, to get 1/36. This is the value of A.

    From this, take x = -3 now:

        \dfrac{2x^5 - 162x + 3}{(x-3)(x^2 + 9)}

    Evaluate and you get -1/36, which is the value of B.

    Now, you are left with:

    \dfrac{2x^5 - 162x + 3}{(x+3)(x-3)(x^2+9)} = \dfrac{1}{36(x - 3)} - \dfrac{1}{36(x + 3)} + \dfrac{Dx + E}{x^2 + 9}

    There are only two constants to find now
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Now, what value do you have to assign to x so that one denominator becomes 0?

    We put x = 3.

    So, you find this:

     \dfrac{2x^5 - 162x + 3}{(x+3)(x^2 + 9)}

    And put x = 3, to get 1/36. This is the value of A.
    but then if you let x = 3, then wouldnt the whole of the LHS = 0? How would this get you a value for A, if you saying \frac{A}{x - 3} = \frac{A}{3 - 3} = 0?
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  4. #4
    MHF Contributor harish21's Avatar
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    look at it this way:

    \dfrac{2x^5 - 162x + 3}{(x-3)(x+3)(x^2+9)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 3} + \dfrac{Dx + E}{x^2 + 9}


    So,

     2x^5 - 162x + 3 = A (x+3)(x^2+9) + B(x-3)(x^2+9) + (Dx+E)(x+3)(x-3)..............................(I)


    Now first put x=-3 in equation (I) to find A.

    then put x=3 in equation (I) to find B..

    Then work on to find D and E
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Yes, that's the same concept, but I find it faster as you don't have to write the full expansion before finding the values of A and B.
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Unknown008 View Post
    Yes, that's the same concept, but I find it faster as you don't have to write the full expansion before finding the values of A and B.
    Thats right! I thought the OP wasn't able to figure out the expansion before finding the values of A and B. So I wrote it down!!
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