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Math Help - Integral - Partial Fraction

  1. #1
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    Integral - Partial Fraction

    Hi I'm trying to solve this Partial Fraction Question: I have found A,B, and C. I just cannot get the correct answer.

    \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx

    A = -\frac{75}{25}
    B = 0
    C = 1

    \int -\frac{\frac{75}{25}}{x-4} + \int \frac{1}{x^2+9}

    Answer:
    -3ln(x-4)+\frac{1}{2}ln(x^2+9)+C
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by killasnake View Post
    Hi I'm trying to solve this Partial Fraction Question: I have found A,B, and C. I just cannot get the correct answer.

    \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx

    A = -\frac{75}{25}
    B = 0
    C = 1

    \int -\frac{\frac{75}{25}}{x-4} + \int \frac{1}{x^2+9}

    Answer:
    -3ln(x-4)+\frac{1}{2}ln(x^2+9)+C
    what were your partial fractions. you do realize that \frac {-3x^2 + x - 31}{(x - 4)(x^2 + 9)} = \frac A{x - 4} + \frac {Bx + C}{x^2 + 9} ?
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  3. #3
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    My Partial Fractions is \frac A{x - 4} + \frac {Bx + C}{x^2 + 9}

     -3x^2+1x-31 = A(x^2+9)+(Bx+C)(x-4)

    Then I plugged in the variables and solved for A,B and C

    <br />
A = -\frac{75}{25}<br />

    <br />
B = 0<br />

    <br />
C = 1<br />

    A,B and C are correct but I cannot get the final correct answer after getting A,B and C
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by killasnake View Post
    My Partial Fractions is \frac A{x - 4} + \frac {Bx + C}{x^2 + 9}

     -3x^2+1x-31 = A(x^2+9)+(Bx+C)(x-4)

    Then I plugged in the variables and solved for A,B and C

    <br />
A = -\frac{75}{25}<br />

    <br />
B = 0<br />

    <br />
C = 1<br />

    A,B and C are correct but I cannot get the final correct answer after getting A,B and C
    that is correct. (why not write A as -3 though?)

    so now you want to find: \int \left( \frac 1{x^2 + 9} - \frac 3{x - 4} <br />
\right)~dx

    the first is an arctan integral  \left( \mbox{we can write it as } \frac 19 \int \frac 1{\left( \frac x3 \right)^2 + 1}~dx \right), the second is a natural log integral. which are you having trouble with?
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  5. #5
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    Well, I solved both intergrals and I believe they are correct. I am just getting frustrated that I am getting the wrong answer.

    <br />
\int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx<br />

    Integral to Solve:
    <br />
\int \left( \frac 1{x^2 + 9} - \frac 3{x - 4} <br />
\right)~dx<br />

    Solved Integral:
    \frac{1}{3} arctan(\frac{1}{3}x) - 3 ln(x-4)
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by killasnake View Post
    Well, I solved both intergrals and I believe they are correct. I am just getting frustrated that I am getting the wrong answer.

    <br />
\int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx<br />

    Integral to Solve:
    <br />
\int \left( \frac 1{x^2 + 9} - \frac 3{x - 4} <br />
\right)~dx<br />

    Solved Integral:
    \frac{1}{3} arctan(\frac{1}{3}x) - 3 ln(x-4)
    well, that's right.

    all's well that ends well
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  7. #7
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    Well at least i know i got the right answer. Thank you so much Jhevon!
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  8. #8
    Math Engineering Student
    Krizalid's Avatar
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    (No partial fractions foundation.)

    x - 3x^2  - 31 = x - 4 - 3\left( {x^2  + 9} \right).
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