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Thread: Integrating fractions, help

  1. #1
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    Integrating fractions, help

    $\displaystyle \int \frac{2x}{3-x} dx $

    I have tired two methods, which both give me wrong answers, here my working , if anyone could tell me where I am going wrong, it would be greatly appreciated.

    method 1: using a substitution:

    $\displaystyle u = 3-x $

    $\displaystyle x = 3-u $

    $\displaystyle dx = -du $


    $\displaystyle \int - \frac{2(3-u)}{u} du $

    $\displaystyle -\int \frac{6-2u}{u} du $ = $\displaystyle -\int \frac{6}{u}-2 \\\ du $ = $\displaystyle - 6ln (3-x) + 2(3-x) + c $


    so the answer is $\displaystyle 6 - 2x -6ln(3-x) +c $

    This is the wrong answer, but I can't see what I have dont wrong, if you anyone could kindly point out.

    Thank you!
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  2. #2
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    Quote Originally Posted by Tweety View Post
    $\displaystyle \int \frac{2x}{3-x} dx $

    I have tired two methods, which both give me wrong answers, here my working , if anyone could tell me where I am going wrong, it would be greatly appreciated.

    method 1: using a substitution:

    $\displaystyle u = 3-x $

    $\displaystyle x = 3-u $

    $\displaystyle dx = -du $

    $\displaystyle \int - \frac{2(3-u)}{u} du $

    $\displaystyle -\int \frac{6-2u}{u} du $ = $\displaystyle -\int \frac{6}{u}-2 \\\ du $ = $\displaystyle - 6ln (3-x) + 2(3-x) + c $


    so the answer is $\displaystyle 6 - 2x -6ln(3-x) +c $

    This is the wrong answer, but I can't see what I have dont wrong, if you anyone could kindly point out.

    Thank you!
    whenever the degree of the numerator is $\displaystyle \ge$ degree of the denominator, perform long division ...

    $\displaystyle \frac{2x}{3-x} = -2 + \frac{6}{3-x}$

    now integrate.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    whenever the degree of the numerator is $\displaystyle \ge$ degree of the denominator, perform long division ...

    $\displaystyle \frac{2x}{3-x} = -2 + \frac{6}{3-x}$

    now integrate.
    thanks,


    Can you also tell me weather this method is valid.

    $\displaystyle \int \frac{4}{2} \times \frac{2}{4} \times \frac{2x}{3-x} $ which is multiplying by one,

    $\displaystyle \frac{4}{2} \int \frac{4x}{-4x+12} $

    = $\displaystyle \frac{4}{2}\int \frac{4x+12-12}{-4x+12} $


    = $\displaystyle \frac{4}{2}\int \frac{4x+12}{-4x+12} - \frac{12}{-4x+12} $ =


    $\displaystyle \frac{4}{2}\int -1 -\frac{12}{-4x+12} $



    $\displaystyle \frac{4}{2} \int -1 - \frac{3}{3-x} $

    which integrates to $\displaystyle \frac{4}{2}( -x - 3ln(3-x) )$

    $\displaystyle -2x - 6ln(3-x) + c $

    This is the correct answer, however I am unsure about weather I am able to multiply by 4/2 and 2/4 and than take 4/2 out of the integral?
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  4. #4
    Member mohammadfawaz's Avatar
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    of course you as long as you are not changing the overall function to integrate. Also $\displaystyle \int_a^b{cf(x)dx} = c\int_a^b{f(x)dx}$ for any constant c.
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