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Thread: Integration of Rational Expression - Alternative Method

  1. #1
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    Integration of Rational Expression - Alternative Method

    Hello everyone,

    I integrated the following rational expression correctly, but one of my steps differ from that one of the book. Could anyone tell me if what I did is still correct?

    Thank you!

    ---

    1. Integrate: $\displaystyle \int \frac{\sqrt{x - 4} + x}{x - 4} dx $

    ---
    $\displaystyle \int \frac{\sqrt{x - 4} + x}{x - 4} dx $

    = $\displaystyle \int (x - 4)^{-1/2} dx + \int \frac{x}{x - 4}dx $

    Let $\displaystyle u = x - 4 $ and $\displaystyle du = dx $. Therefore, $\displaystyle x = u + 4$.

    = $\displaystyle \int u^{-1/2} du + \int \frac{u + 4}{u}du $

    = $\displaystyle 2\sqrt{x - 4} + \int \frac{u}{u} du + \int \frac{4}{u} du$

    = $\displaystyle 2\sqrt{x - 4} + x + \ln(x - 4)^4 + C$

    ---

    However, my practice workbook integrated $\displaystyle \int \frac{u}{u} du $ as $\displaystyle u $, and then substituted $\displaystyle x - 4 $ back into the equation.

    So according to the workbook, its last step would be:

    = $\displaystyle 2\sqrt{x - 4} + u + \ln(x - 4)^4 + C$

    = $\displaystyle 2\sqrt{x - 4} + (x - 4) + \ln(x - 4)^4 + C$

    = $\displaystyle 2\sqrt{x - 4} + x + \ln(x - 4)^4 + C$
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  2. #2
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    I think the way you did it kind of implies that $\displaystyle \int \frac{u}{u}du=x$ because of the way you kept your terms in order. The book's method is more correct in my opinion as it notes that the above integral doesn't equal x, but u + C, then back-substitutes x-4 for u and combines all the constants into one C.
    Last edited by Jameson; Aug 5th 2008 at 06:27 AM. Reason: small mistake
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  3. #3
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    Hello, scherz0!

    What you did is correct.
    . . I would not have simplifed first as you did . . .


    We are given: .$\displaystyle \int \frac{\sqrt{x - 4} + x}{x - 4}\,dx $

    Let: $\displaystyle \sqrt{x-4} \:=\: u \quad\Rightarrow\quad x-4\:=\:u^2 \quad\Rightarrow\quad x \:=\: u^2+4 \quad\Rightarrow\quad dx \:=\:2u\,du$


    Substitute: .$\displaystyle \int \frac{u + (u^2+4)}{u^2}\,(2u\,du) \;\;=\;\;2\int\frac{u^2+u-4}{u}\,du \;\;=\;\; 2\int\left(u + 1 - \frac{4}{u}\right)\,du $

    . . . . . . $\displaystyle = \;\;2\left(\frac{u^2}{2} + u - 4\ln|u|\right) + C \;\;=\;\;u^2 + 2u - 8 \ln|u| + C$


    Back-substitute: .$\displaystyle (x - 4) + 2\sqrt{x-4} - 8\ln\left(\sqrt{x-4}\right) + C $

    . . . . . . . . . . .$\displaystyle =\;\;x - 4 + 2\sqrt{x-4} - 8\ln(x-4)^{\frac{1}{2}} + C$

    . . . . . . . . . . .$\displaystyle = \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + \underbrace{C - 4}_{\text{a constant}} $

    . . . . . . . . . . .$\displaystyle = \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + C$

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