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Math Help - More partial fractions fun

  1. #1
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    More partial fractions fun

    in the case of INTEGRAL OF: (x^3 + 7) / (x^2 + 4)

    If you do long division of polynomials you get:

    INTEGRAL OF: X + (-4x + 7) / (x^2 + 4)

    I tried doing partial fractions here, but it doesn't work. The problem i run into is that i only have one term in my denominator, it cannot be broken up. Therefore i cant do any of the 4 cases.

    I thought of something new:

    I broke the fraction down into:
    (-4x) / (x^2 + 4) and (7) / (x^2 + 4)

    The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
    I determined that the integral of that was 7( -x^[-1] + .25x)
    Then i set this aside for later, knowing I will eventually have to add it to my other integral.

    Next i looked at (-4x) / (x^2 + 4)
    I pulled a -2 out and got then it looked like:
    -2 INTEGRAL OF: ( 2x / ((x^2) + 4)
    I decided to do u substitution for u = x^2 + 4 and my du was just 2xdx making dx=du/2x
    So when i plug in u i get:

    -2 INTEGRAL of: (2x / u) * (1/2x)du
    The 2x's cancel out and your left with -2 *INTEGRAL of: 1/u du
    This is clearly -2ln(u). I plug back in u and get -2ln(x^2 + 4)

    So my final answer is .5x^2 - 2ln(x^2 + 4) - 7/x + 7/4x

    But this is wrong, or so says my book.
    Last edited by fogel1497; September 26th 2008 at 03:14 AM.
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  2. #2
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    Quote Originally Posted by fogel1497 View Post
    in the case of INTEGRAL OF: (x^3 + 7) / (x^2 + 4)

    If you do long division of polynomials you get:

    INTEGRAL OF: X + (-4x + 7) / (x^2 + 4)

    I tried doing partial fractions here, but it doesn't work. The problem i run into is that i only have one term in my denominator, it cannot be broken up. Therefore i cant do any of the 4 cases.

    I thought of something new:

    I broke the fraction down into:
    (-4x) / (x^2 + 4) and (7) / (x^2 + 4) Mr F says: Good.

    The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
    I determined that the integral of that was 7( -x^[-1] + .25x) Mr F says: Not so good. See below.

    Then i set this aside for later, knowing I will eventually have to add it to my other integral.

    Next i looked at (-4x) / (x^2 + 4)
    I pulled a -2 out and got then it looked like:
    -2 INTEGRAL OF: ( 2x / ((x^2) + 4)
    I decided to do u substitution for u = x^2 + 4 and my du was just 2xdx making dx=du/2x
    So when i plug in u i get:

    -2 INTEGRAL of: (2x / u) * (1/2x)du
    The 2x's cancel out and your left with -2 *INTEGRAL of: 1/u du
    This is clearly -2ln(u). I plug back in u and get -2ln(x^2 + 4) Mr F says: Good.

    So my final answer is .5x^2 - 2ln(x^2 + 4) - 7/x + 7/4x Mr F says: Not so good.

    But this is wrong, or so says my book.
    Standard form: \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C
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  3. #3
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    Quote Originally Posted by fogel1497 View Post
    The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
    I determined that the integral of that was 7( -x^[-1] + .25x)
    Then i set this aside for later, knowing I will eventually have to add it to my other integral.
    This is where you went wrong. Firstly:
    \frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}

    \frac{1}{x^2 + 4} \neq x^{-2} + \frac{1}{4}

    What you should do here is pull 4 common factor to get:

    \frac{7}{4} \int \frac{1}{(\frac{x}{2})^2 + 1}

    You should recognize this right away as a standard form (arctan).
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  4. #4
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    x^2+4) = x-\frac{4x-7}{x^2+4}
    " alt="
    (x^3+7)x^2+4) = x-\frac{4x-7}{x^2+4}
    " />
     <br />
\int{xdx} - 4\int{\frac{xdx}{x^2+4}} + 7\int{\frac{dx}{x^2+4}}<br />

    and the answer is
     <br />
\frac{x^2}{2}-2\ln{|x^2+4|}+\frac{7}{2}\arctan{\frac{x}{2}}+c<br />
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