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Thread: 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

    $\displaystyle \frac{3x^2-2x+12}{(x^2+4)^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{( x^2+4)^2}$
    Multiply through by the denominator,
    $\displaystyle 3x^2-2x+12=(Ax+B)(x^2+4)+(Cx+D)$
    $\displaystyle 3x^2-2x+12=Ax^3+Bx^2+4Ax+4B+Cx+D$
    $\displaystyle 3x^2-2x+12=Ax^3+Bx^2+(4A+C)x+(4B+D)$
    Our eyes speaketh to us and say,
    $\displaystyle A=0$
    $\displaystyle B=3$
    Thus,
    $\displaystyle 4A+C=4(0)+C=-2 \to C=-2$
    $\displaystyle 4B+D=4(3)+D=12 \to D=0$
    Thus,
    $\displaystyle \frac{3}{x^2+4}-\frac{2x}{(x^2+2)^2}$
    The first integral,
    $\displaystyle \int \frac{3}{x^2+4} dx = \frac{3}{4} \int \frac{1}{\frac{x^2}{4}+1} dx=\frac{3}{4} \int \frac{1}{1+(x/2)^2} dx = \frac{3}{4} \cdot 2 \cdot \tan^{-1} \frac{x}{2}+C$
    The second integral,
    $\displaystyle \int \frac{2x}{(x^2+2)^2}dx$
    Let $\displaystyle u=x^2+2$
    Thus,
    $\displaystyle \int \frac{1}{u^2} du$
    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,
    $\displaystyle \int \frac{2x}{x^2-2x+2} dx$
    Add an subtract,
    $\displaystyle \int \frac{2x-2+2}{x^2-2x+2} dx$
    Thus,
    $\displaystyle \int \frac{2x-2}{x^2-2x+2} dx + \int \frac{2}{x^2-2x+2} dx$
    The first integral is trivial, use $\displaystyle u=x^2-2x+2$.
    Thus, we get (details omitted),
    $\displaystyle \ln |x^2-2x+2|+C$.

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

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

    Only the first one requires partial fractions . . .


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

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

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

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


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


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

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


    $\displaystyle \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}
    \begin{array}{cc}\Rightarrow \\ \Rightarrow \end{array}
    \begin{array}{cc}5A + C \:=\:\text{-}2 \\ 8A + C\:=\:\text{-}2\end{array}
    \begin{array}{cc}(5)\\(6)\end{array}
    $

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


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

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


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