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Math Help - not my favourite area:

  1. #1
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    not my favourite area:

    im not the biggest fan of partial fraction, doing so practice excercise to get me in the swing of things before going back to uni and have stumbled across this:

    f(x) = 3
    x^3 +3x^2 -10x

    1) write f(x) as a sum of partial fractions
    2) evaluate (intergrate f(x))
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  2. #2
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    Start by factoring the denominator.

    x^{3}+3x^{2}-10x=x(x-2)(x+5)

    That gives:

    \frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+5}=3

    Now, can you solve for A,B,C?.
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  3. #3
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    is this correct so far:

    3= A(x+5) + B(x-2)(x) + C(x)

    amanda x
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    is this correct so far:

    3= A(x+5) + B(x-2)(x) + C(x)

    amanda x
    Almost. Take it step by step and add the fractions:

    \frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+5}=\frac{3}{x(x - 2)(x + 5)}

    \frac{A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2)}{x(x - 2)(x + 5)} = \frac{3}{x(x - 2)(x + 5)}

    So
    A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2) = 3

    -Dan
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  5. #5
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    thank you both of you. trying to get my head around intergrating it now :s
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  6. #6
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    i have managed to get
    b = 3/15 (by using 2)
    c = 3/35 by using -5)

    cant find the number to use to work out a :s
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Almost. Take it step by step and add the fractions:

    \frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+5}=\frac{3}{x(x - 2)(x + 5)}

    \frac{A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2)}{x(x - 2)(x + 5)} = \frac{3}{x(x - 2)(x + 5)}

    So
    A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2) = 3

    -Dan
    Allow me to continue, then.

    A(x^2 + 3x - 10) + B(x^2 + 5x) + C(x^2 - 2x) = 3

    (A + B + C)x^2 + (3A + 5B - 2C)x - 10A = 3

    Thus
    A + B + C = 0
    3A + 5B - 2C = 0
    -10A = 3

    So
    A = -\frac{3}{10}

    This implies the other two equations are:
    B + C = \frac{3}{10}
    5B - 2C = \frac{9}{10}

    From the first equation:
    C = \frac{3}{10} - B

    Inserting this into the second equation:
    5B - 2 \left ( \frac{3}{10} - B \right ) = \frac{9}{10}

    B = \frac{3}{14}

    C = \frac{3}{10} - \frac{3}{14} = \frac{3}{35}

    -Dan
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  8. #8
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    Hello, Amanda!

    f(x) \:=\:\frac{3}{x^3+3x^2 - 10x}

    (1) Write f(x) as a sum of partial fractions

    (2) Find: . \int f(x)\,dx
    You seem to know the "fast" way for Partial Fractions.
    I find it much easier, but many teachers don't know it
    . . and others actively disapprove of it.

    We have: . \frac{3}{x(x-2)(x+5)} \:=\:\frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+5}

    Multiply through the LCD: . 3 \;=\;(x-2)(x+5)A + x(x+5)B + x(x-2)C


    Now select some "nice" values for x. .
    (You'll see what I mean)

    . . Let x=5\!:\;\;3 \;=\;(\text{-}2)(5)A + 0\!\cdot\!B + 0\!\cdot\!C\quad\Rightarrow\quad A \:=\:\text{-}\frac{3}{10}

    . . Let x = 2\!:\;\;3 \;=\;0\!\cdot\!A + 2\!\cdot\!7\!\cdot\!B + 0\!\cdot\!C \quad\Rightarrow\quad B \:=\:\frac{3}{14}

    . . Let {\color{blue}x = \text{-}5}\!:\;\;3 \;=\;0\!\cdot\!A + 0\!\cdot\!B + (\text{-}5)(\text{-}7)C\quad\Rightarrow\quad C \:=\:\frac{3}{35}

    And we have: . \frac{-\frac{3}{10}}{x} + \frac{\frac{3}{14}}{x-2} + \frac{\frac{3}{35}}{x+5} \;=\;\frac{3}{70}\left[-\frac{7}{x} + \frac{5}{x-2} + \frac{2}{x+5}\right] . (1)



    Integrate: . \frac{3}{70}\int\left[-\frac{7}{x} + \frac{5}{x-2} + \frac{2}{x+5}\right]\,dx \;= \;\frac{3}{70}\bigg[-7\ln|x|\, +\, 5\ln|x-2|\, +\, 2\ln|x+5|\bigg]\,+\,C . (2)



    If you want to show-off, you can simplify further . . .

    \frac{3}{70}\bigg[-\ln|x^7| + \ln|x-2|^5) + \ln(x+5)^2\bigg] + C \;=\; \frac{3}{70}\ln\left|\frac{(x-2)^5(x+5)^2}{x^7}\right| + C

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