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Math Help - Partial fractions, indefinte integral

  1. #1
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    Arrow Partial fractions, indefinte integral

    Hi, I have the problem:

    and it states that the integrand has partial fractions decomposition:

    So then I need to make 7x^3+5x^2+125x+75=a(x)(x^2+25)+b(x^2)(cx+d)+(cx+d)  (x^3)
    Now, would ax^3=7? I know what to do with simple A and B partial fractions but this is looking difficult especially when there is multiple variables multiplied by each other. How can I solve this?
    Thanks,
    Matt
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by matt3D View Post
    Hi, I have the problem:

    and it states that the integrand has partial fractions decomposition:

    So then I need to make 7x^3+5x^2+125x+75=a(x)(x^2+25)+b(x^2)(cx+d)+(cx+d)  (x^3)
    Now, would ax^3=7? I know what to do with simple A and B partial fractions but this is looking difficult especially when there is multiple variables multiplied by each other. How can I solve this?
    Thanks,
    Matt
    Nope, you have to develop completely (and correctly lol)

    \frac{a}{x^2}+\frac bx+\frac{cx+d}{x^2+25}
    The common denominator is x^2(x^2+25)

    \implies =\frac{a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)}{x^2(x^2  +25)}

    Please spot out your mistakes...


    a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)=\dots=x^3(b+c)+  x^2(a+d)+x(25b)+25(a)
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  3. #3
    Super Member 11rdc11's Avatar
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    A=3
    B=5
    C=2
    D=2
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  4. #4
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    Hello, matt3D!

    \int\frac{7x^3 + 5x^2 + 125x + 75}{x^4+25x^2}\,dx

    We have: . \frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{Cx}{x^2+25} + \frac{D}{x^2+25}


    Note that I immediately made two fractions for the quadratic.
    I discovered this trick a half-century ago and have taught it that way ever since.
    It seems that no teacher has discovered this yet!


    Multiply through by the LCD:

    . . 7x^3 + 5x^2 + 125x + 75 \;=\;x(x^2+16)\!\cdot\!A + (x^2+25)\!\cdot\!B + x^3\!\cdot\!C + x^2\!\cdot\!D


    Let x = 0\!:\quad 75 \;=\;25B \quad\Rightarrow\quad\boxed{B \:=\:3}


    Let x = 1\!:\quad 212 \:=\:26A + 26B + C + D \quad\Rightarrow\quad 26A + C + D \:=\:134\;\;{\color{blue}[1]}


    Let x = \text{-}1\!:\quad \text{-}52 \:=\:-26A + 26B - C + D \quad\Rightarrow\quad \text{-}26A - C + D \:=\:\text{-}130\;\;{\color{blue}[2]}


    Let x = 2\!:\quad 401 \:=\:58A + 29B + 8C + 4D \quad\Rightarrow\quad29A + 4C + 2D \:=\:157\;\;{\color{blue}[3]}


    Add [1] and [2]: . 2D \:=\:4 \quad\Rightarrow\quad\boxed{ D \:=\:2}


    \begin{array}{cccc}\text{Then {\color{blue}[1]} becomes:} & 26A + C &=& 132 \\<br />
\text{and {\color{blue}[3]} becomes:} & 29A + 4C &=& 153 \end{array}

    Solve this system: . \boxed{A \:=\:5}\quad\boxed{ C \:=\:2}


    Therefore: . \frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{5}{x} + \frac{3}{x^2} + \frac{2x}{x^2+25} + \frac{2}{x^2+25}

    . . which can be readily integrated . . .

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  5. #5
    Moo
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    It seems that no teacher has discovered this yet!
    Well, my teacher taught it to us... and I'm sure he's not the only one. Some people used it on MHF too

    And this teacher went further than this by taking the derivative when we had something like x and x in the partial fractions decomposition. ^^'
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  6. #6
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    Quote Originally Posted by Moo View Post
    Hello,

    Nope, you have to develop completely (and correctly lol)

    \frac{a}{x^2}+\frac bx+\frac{cx+d}{x^2+25}
    The common denominator is x^2(x^2+25)

    \implies =\frac{a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)}{x^2(x^2  +25)}

    Please spot out your mistakes...


    a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)=\dots=x^3(b+c)+  x^2(a+d)+x(25b)+25(a)
    Yea, I was super tired when I was doing this.
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, matt3D!


    We have: . \frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{Cx}{x^2+25} + \frac{D}{x^2+25}


    Note that I immediately made two fractions for the quadratic.
    I discovered this trick a half-century ago and have taught it that way ever since.
    It seems that no teacher has discovered this yet!


    Multiply through by the LCD:

    . . 7x^3 + 5x^2 + 125x + 75 \;=\;x(x^2+16)\!\cdot\!A + (x^2+25)\!\cdot\!B + x^3\!\cdot\!C + x^2\!\cdot\!D


    Let x = 0\!:\quad 75 \;=\;25B \quad\Rightarrow\quad\boxed{B \:=\:3}


    Let x = 1\!:\quad 212 \:=\:26A + 26B + C + D \quad\Rightarrow\quad 26A + C + D \:=\:134\;\;{\color{blue}[1]}


    Let x = \text{-}1\!:\quad \text{-}52 \:=\:-26A + 26B - C + D \quad\Rightarrow\quad \text{-}26A - C + D \:=\:\text{-}130\;\;{\color{blue}[2]}


    Let x = 2\!:\quad 401 \:=\:58A + 29B + 8C + 4D \quad\Rightarrow\quad29A + 4C + 2D \:=\:157\;\;{\color{blue}[3]}


    Add [1] and [2]: . 2D \:=\:4 \quad\Rightarrow\quad\boxed{ D \:=\:2}


    \begin{array}{cccc}\text{Then {\color{blue}[1]} becomes:} & 26A + C &=& 132 \\<br />
\text{and {\color{blue}[3]} becomes:} & 29A + 4C &=& 153 \end{array}

    Solve this system: . \boxed{A \:=\:5}\quad\boxed{ C \:=\:2}


    Therefore: . \frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{5}{x} + \frac{3}{x^2} + \frac{2x}{x^2+25} + \frac{2}{x^2+25}

    . . which can be readily integrated . . .

    Hi Soroban, thanks for the explanation. But why do you multiply A by x(x^2+16) I don't see that in the fractions, and D by x^2 and C by x^3?
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