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Math Help - partial fractions

  1. #1
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    partial fractions

    I'm getting stuck on these, so if someone can please help, I'd really appreciate it.

    Use partial fractions to evaluate the integral.

    1. ∫(3xsquared - 2x + 12) / (xsquared + 4) squared

    2. ∫ (2r) / (rsquared - 2r + 2)

    3.∫ (theta) / (theta + 1)

    Thanks!
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  2. #2
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    Quote Originally Posted by turtle View Post
    I'm getting stuck on these, so if someone can please help, I'd really appreciate it.

    Use partial fractions to evaluate the integral.

    1. ∫(3xsquared - 2x + 12) / (xsquared + 4) squared

    \frac{3x^2-2x+12}{(x^2+4)^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(  x^2+4)^2}
    Multiply through by the denominator,
    3x^2-2x+12=(Ax+B)(x^2+4)+(Cx+D)
    3x^2-2x+12=Ax^3+Bx^2+4Ax+4B+Cx+D
    3x^2-2x+12=Ax^3+Bx^2+(4A+C)x+(4B+D)
    Our eyes speaketh to us and say,
    A=0
    B=3
    Thus,
    4A+C=4(0)+C=-2 \to C=-2
    4B+D=4(3)+D=12 \to D=0
    Thus,
    \frac{3}{x^2+4}-\frac{2x}{(x^2+2)^2}
    The first integral,
    \int \frac{3}{x^2+4} dx = \frac{3}{4} \int \frac{1}{\frac{x^2}{4}+1} dx=\frac{3}{4} \int \frac{1}{1+(x/2)^2} dx = \frac{3}{4} \cdot 2 \cdot \tan^{-1} \frac{x}{2}+C
    The second integral,
    \int \frac{2x}{(x^2+2)^2}dx
    Let u=x^2+2
    Thus,
    \int \frac{1}{u^2} du
    The rest is trivial.
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  3. #3
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    Quote Originally Posted by turtle View Post

    2. ∫ (2r) / (rsquared - 2r + 2)
    You have,
    \int \frac{2x}{x^2-2x+2} dx
    Add an subtract,
    \int \frac{2x-2+2}{x^2-2x+2} dx
    Thus,
    \int \frac{2x-2}{x^2-2x+2} dx + \int \frac{2}{x^2-2x+2} dx
    The first integral is trivial, use u=x^2-2x+2.
    Thus, we get (details omitted),
     \ln |x^2-2x+2|+C.

    The second integral needs some work, first we have,
    \int \frac{2}{x^2-2x+1+1} dx
    Complete the square,
    \int \frac{2}{(x-1)^2+1} dx
    The integral is trivial, use u=x-1.
    Thus,
    2\tan^{-1}(x-1)+C
    Thus,
    \ln |x^2-2x+2|+2\tan^{-1}(x-1)+C
    Note you can write,
    \ln (x^2-2x+2)+2\tan^{-1}(x-1)+C
    Becuase,
    x^2-2x+2>0
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  4. #4
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    Quote Originally Posted by turtle View Post

    3.∫ (theta) / (theta + 1)
    Add and subtract,
    \int \frac{x+1-1}{x+1} dx
    Thus,
    \int 1 dx - \int \frac{1}{x+1} dx
    The first integral is trivial,
    x+C
    The second integral is trivial, use u=x+1.
    Thus,
    x+\ln |x+1|+C
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  5. #5
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    Hello, turtle!

    Only the first one requires partial fractions . . .


    1)\;\int\frac{3x^2-2x+12}{(x^2+4)^2}\,dx

    We have: . \frac{3x^2-2x+12}{(x^2+4)^2} \;=\;\frac{Ax + B}{x^2 + 4} + \frac{Cx+d}{(x^2+4)^2}

    Clear denominators: . 3x^2-2x+2\;=\;Ax(x^2+4) + B(x^2+4) + Cx + D

    \begin{array}{cccc}\text{Let }x = 0\!: & 12\;=\;A\cdot0 + B\cdot4 + C\cdot0 + D  \\\text{Let }x = 1\!: & 13\;=\;A\cdot5 + B\cdot5+ C + D \\ \text{Let }x = \text{-}1\!: & 17 \;=\;A(\text{-}5) + B\cdot5 - C + D \\ <br />
\text{Let }x = 2\!: & 20 \;=\;A\cdot16 + B\cdot8 + C\cdot2 + D \end{array}


    Solve the system of equations: . \begin{array}{cccc} 4B + D \\ 5A + 5B + C + D \\ -5A + 5B - C + D \\ 16A + 8B + 2C + D\end{array} \begin{array}{cccc}=\\=\\=\\=\end{array} \begin{array}{cccc}12\\13\\17\\20\end{array}\; \begin{array}{cccc}(1)\\(2)\\(3)\\(4)\end{array}


    Add (2) and (3): . 10B + 2D\:=\:30\quad\Rightarrow\quad 5B + D \:=\:15
    . . . Subtract (1): . . . . . . . . . . . . . . . . . 4B + D \:=\:12

    We get: . \boxed{B = 3} . . . hence: . \boxed{D = 0}


    \begin{array}{cc}\text{Substitute into (2):} & 5A + 5(3) + C + 0 \:=\:13 \\ \text{Substitute into (4):} & 16A + 8(3) + 2C + 0 \:=\:20 \end{array}<br />
\begin{array}{cc}\Rightarrow \\ \Rightarrow \end{array}<br />
\begin{array}{cc}5A + C \:=\:\text{-}2 \\ 8A + C\:=\:\text{-}2\end{array}<br />
\begin{array}{cc}(5)\\(6)\end{array}<br />

    Subtract (5) from (6): . 3A \:=\:0\quad\Rightarrow\quad\boxed{ A = 0} . . . hence: . \boxed{C = \text{-}2}


    The integral becomes: . \int\left(\frac{3}{x^2+4} - \frac{2x}{(x^2+4)^2}\right)dx

    The first is of the \arctan form.
    For the second, let u \:=\:x^2+4


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