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Math Help - Partial Fractions -- any tricks or shortcuts?

  1. #1
    Junior Member cinder's Avatar
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    Partial Fractions -- any tricks or shortcuts?

    Today in class we did partial fractions. I understand the process, but I was wondering if there are any shortcuts.

    For example: perhaps a shortcut way to go from \int\frac{x^2+5}{x^3-x^2+x+3}dx to \int\frac{2}{x^2-2x+3} + \frac{1}{x+1}dx without all the Ax+B stuff in between.
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  2. #2
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    Hello, cinder!

    I was wondering if there are any shortcuts.

    Perhaps a way to go from \int\frac{x^2+5}{x^3-x^2+x+3}\,dx .to . \int\left(\frac{2}{x^2-2x+3} + \frac{1}{x+1}\right)dx

    . . without all the Ax+B stuff in between.
    With a prime quadratic factor, there's no way to avoid the Ax + B.
    . . But there are ways to make life a bit easier.


    We have: . \frac{x^2 + 5}{(x+1)(x^2-2x+3)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2-2x+3} + \frac{C}{x^2-2x+3}

    . . I immediately make two fractions for the quadratic.


    Clear denominators: . x^2 + 5 \;=\;A(x^2-2x+3) + Bx(x+1) + C(x + 1)


    Let x = \text{-}1\!:\;\;6\;=\;A(6) + B(0) + C(0)\quad\Rightarrow\quad\boxed{A = 1}

    . . I immediately replace A with 1\!:\;\;x^2 + 5 \;=\;(x^2 - 2x + 3) + Bx(x + 1) + C(x + 1)


    Let x = 0\!:\;\;5 \;=\;(3) + B(0) + C(1)\quad\Rightarrow\quad\boxed{ C = 2}

    . . Replace C with 2\!:\;\;x^2 + 5\;=\;(x^2-2x+3) + Bx(x+1) + 2(x+1)


    Let x = 1\!:\;\;6 \;=\;(2) + B(2) + (4)\quad\Rightarrow\quad\boxed{B = 0}


    Therefore: . \frac{x^2+5}{(x+1)(x^2-2x+3)} \;=\;\frac{1}{x+1} + \frac{2}{x^2-2x+3}

    . . . . . with a minimum of pain and aggravation.

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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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     <br />
\begin{aligned}<br />
x^3 - x^2 + x + 3 &= x^3 - (2x^2 - x^2 ) + (3x - 2x) + 3\\<br />
&= (x^3 + x^2 ) - (2x^2 + 2x) + (3x + 3)\\<br />
&= x^2 (x + 1) - 2x(x + 1) + 3(x + 1)\\<br />
&= (x + 1)(x^2 - 2x + 3)<br />
\end{aligned}<br />

    Therefore

    \begin{aligned}<br />
\int {\frac{{x^2 + 5}}<br />
{{x^3 - x^2 + x + 3}}~dx} &= \int {\frac{{x^2 + 5}}<br />
{{(x + 1)(x^2 - 2x + 3)}}~dx}\\<br />
&= \int {\frac{{2(x + 1) + (x^2 - 2x + 3)}}<br />
{{(x + 1)(x^2 - 2x + 3)}}~dx}\\<br />
&= \int {\left( {\frac{2}<br />
{{x^2 - 2x + 3}} + \frac{1}<br />
{{x + 1}}} \right)~dx}<br />
\end{aligned}
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