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Math Help - Partial Fraction Quadratic Factor Integration

  1. #1
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    Partial Fraction Quadratic Factor Integration

    I think I understand the basic idea behind partial fractions but I am stuck on a couple of problems with quadratic factors. I think if I saw one done I could figure out the rest. This is one of the ones giving me trouble. When I am done with the partial fraction I still seem unable to integrate it. I'm about a month behind in my online class and already have dropped it once so am hoping not to have to again.

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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CrazyLond View Post
    I think I understand the basic idea behind partial fractions but I am stuck on a couple of problems with quadratic factors. I think if I saw one done I could figure out the rest. This is one of the ones giving me trouble. When I am done with the partial fraction I still seem unable to integrate it. I'm about a month behind in my online class and already have dropped it once so am hoping not to have to again.
    Setting up our partial fractions we get



    \frac{x^2-x+9}{(x^2+9)^2}=\frac{Ax+B}{x^2+9}+\frac{Cx+D}{(x^  2+9)^2}

    Multiplying through by (x^2+9)^2

    we get

    x^2-x+9=(Ax+B)(x^2+9)+Cx+D=Ax^3+Bx^2+9Ax+Cx+9B+D

    Grouping we get x^2-x+9=Ax^3+Bx^2+(9A+c)x+9B+D

    Equating Coefficients we first get

    Ax^3=0x^3\Rightarrow{A=0}

    Next we have that Bx^2=x^2\Rightarrow{B=1}

    Next we have 9A+c=9\cdot{0}+C=C=-1

    And finally 9B+D=9+D=9\Rightarrow{D=0}

    Putting it back into our original we get

    \frac{x^2-x+9}{(x^2+9)^2}=\frac{1}{x^2+9}+\frac{-x}{(x^2+9)^2}

    So now \int\frac{x^2-x+9}{(x^2+9)^2}dx=\int\bigg[\frac{1}{x^2+9}-\frac{x}{(x^2+<br />
9)^2}\bigg]dx

    The first integral is a arctangent so

    \int\frac{dx}{(3)^2+x^2}=\frac{1}{3}arctan\bigg(\f  rac{x}{3}\bigg)+C

    Noticing in the second one that x is 1/2 the derivative of the quantity

    we rewrite to get

    \int\frac{x}{(x^2+9)^2}=\frac{1}{2}\int\frac{2x}{(  x^2+9)^2}dx=\frac{-1}{2(x^2+9)}+C

    Combining we get

    \int\frac{x^2-x+9}{(x^2+9)^2}dx=\frac{1}{3}arctan\bigg(\frac{x}{  3}\bigg)+\frac{1}{2(x^2+9)}+C
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  3. #3
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    Quote Originally Posted by CrazyLond View Post
    I think I understand the basic idea behind partial fractions but I am stuck on a couple of problems with quadratic factors. I think if I saw one done I could figure out the rest. This is one of the ones giving me trouble. When I am done with the partial fraction I still seem unable to integrate it. I'm about a month behind in my online class and already have dropped it once so am hoping not to have to again.

    No partial fractions are needed rewrite as
    \frac{x^2+9-x}{(x^2+9)^2}=\frac{x^2+9}{(x^2+9)^2}+\frac{-x}{(x^2+9)^2}=\frac{1}{x^2+9}-\frac{x}{(x^2+9)^2}
    Last edited by TheEmptySet; May 16th 2008 at 08:18 PM. Reason: typo
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    No partial fractions are needed rewrite as
    \frac{x^2+9-x}{(x^2+9)^2}=\frac{x^2+9}{(x^2+9)^2}+\frac{-x}{(x^+9)^2}
    Yeah but wasn't he asking for partial fractions decomposition?
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Mathstud28 View Post
    Yeah but wasn't he asking for partial fractions decomposition?
    Yes, but I think it is good for poeple to see other methods.

    I didn't start "seeing" these for quite a long time, and I still miss them now.

    This is just my opinion.

    What do you think?

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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Yes, but I think it is good for poeple to see other methods.

    I didn't start "seeing" these for quite a long time, and I still miss them now.

    This is just my opinion.

    What do you think?

    Well...I agree with you whole-heartdly...I was even about to do it that way...but I get chastised on here all the time for not answering what the poster directly asked....so I didnt

    All in all I would say both ways are worth seeing
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  7. #7
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    Thanks for the help. My problem was multiplying A and B by the square of what I should have.

    The problem did ask for me to use partial fractions but thanks for showing that other method.
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