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Math Help - Another Integration by Partial Fractions

  1. #1
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    Another Integration by Partial Fractions

    Hello,

    \int\frac{-x^2-x-1}{5x^2+7x-6}dx =


    <br />
\int[\frac{A}{5x-3} + \frac{B}{x+2}]dx

    and

    <br />
B(5x-3) + A(x+2) = -x^2-x-1

    I'm having trouble solving A and B. Is there a method for determining them? Or should I be approaching the problem differently altogether?

    Thanks for your consideration,

    Austin Martin
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  2. #2
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    Quote Originally Posted by auslmar View Post
    Hello,

    \int\frac{-x^2-x-1}{5x^2+7x-6}dx =


    <br />
\int[\frac{A}{5x-3} + \frac{B}{x+2}]dx

    and

    <br />
B(5x-3) + A(x+2) = -x^2-x-1

    I'm having trouble solving A and B. Is there a method for determining them? Or should I be approaching the problem differently altogether?

    Thanks for your consideration,

    Austin Martin
    Since the degree of the numberator and the denominator are the same you need to use long division before partial fractions!

    Remember that if the degree of the numerator is greater than or equal to the denominator ALWAYS use long division.

    I don't know how to write out long division with LaTex but

    \frac{-x^2-x-1}{5x^2+7x-6}=-\frac{1}{5}+\frac{\frac{2}{5}x-\frac{11}{5}}{5x^2+7x-6}=-\frac{1}{5}+\frac{1}{5}\left( \frac{2x- 11}{5x^2+7x-6} \right)

    Now we can use partial fractions on \left( \frac{2x- 11}{5x^2+7x-6} \right)

    Good luck
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    Since the degree of the numberator and the denominator are the same you need to use long division before partial fractions!

    Remember that if the degree of the numerator is greater than or equal to the denominator ALWAYS use long division.

    I don't know how to write out long division with LaTex but

    \frac{-x^2-x-1}{5x^2+7x-6}=-\frac{1}{5}+\frac{\frac{2}{5}x-\frac{11}{5}}{5x^2+7x-6}=-\frac{1}{5}+\frac{1}{5}\left( \frac{2x- 11}{5x^2+7x-6} \right)

    Now we can use partial fractions on \left( \frac{2x- 11}{5x^2+7x-6} \right)

    Good luck
    Ah yes! I completely forgot about the long division bit. Thanks for the help . I'll work it out!
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