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Math Help - Integrating fractions, help

  1. #1
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    Integrating fractions, help

     \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:

     u = 3-x

     x = 3-u

     dx = -du


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

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


    so the answer is  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
     \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:

     u = 3-x

     x = 3-u

     dx = -du

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

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


    so the answer is  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 \ge degree of the denominator, perform long division ...

    \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 \ge degree of the denominator, perform long division ...

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

    now integrate.
    thanks,


    Can you also tell me weather this method is valid.

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

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

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


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


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



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

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

     -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 \int_a^b{cf(x)dx} = c\int_a^b{f(x)dx} for any constant c.
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