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Math Help - Integration problem help... partial fraction...

  1. #1
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    Integration problem help... partial fraction...

    I can't seem to figure out this partial fractions problem:

    \int \frac {dx}{3x-3x^2}

    My work so far has involved factoring out a 3x in the denominator so the problem then looks like \int \frac {1}{3x(1-x)}, and then multiplying through using the cover up rule, but so far that method hasn't resulted in the answer:

    (\frac {1}{3})\ln \frac {x}{(1-x)} + c
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  2. #2
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    Did you find A and B where \frac {1}{3x(1-x)} = \frac {A}{3x} +\frac {B}{1-x} ?
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    Quote Originally Posted by sgcb View Post
    I can't seem to figure out this partial fractions problem:

    \int \frac {dx}{3x-3x^2}

    My work so far has involved factoring out a 3x in the denominator so the problem then looks like \int \frac {1}{3x(1-x)}, and then multiplying through using the cover up rule, but so far that method hasn't resulted in the answer:

    (\frac {1}{3})\ln \frac {x}{(1-x)} + c
    I'm not sure what rule your referring to. I would just decompose the function.

    \int\frac{1}{3x(1-x)}dx=\int\left(\frac{A}{3x}+\frac{B}{1-x}\right)dx

    Are you familiar with this method?

    A(1-x)+B(3x)=1

    x(3B-A)+A=1

    So now you solve the system of equations:
    --------------------------

    3B-A=0

    A=1
    --------------------------

    Substitute those values for A and B into the integral, and it's pretty easy from there.
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  4. #4
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    Quote Originally Posted by sgcb View Post
    I can't seem to figure out this partial fractions problem:

    \int \frac {dx}{3x-3x^2}

    My work so far has involved factoring out a 3x in the denominator so the problem then looks like \int \frac {1}{3x(1-x)}, and then multiplying through using the cover up rule, but so far that method hasn't resulted in the answer:

    (\frac {1}{3})\ln \frac {x}{(1-x)} + c
    note that \frac{1}{3x(1-x)} = \frac{1}{3} \cdot \frac{1}{x(1-x)}

    and ...

    \frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x}<br />
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  5. #5
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    Quote Originally Posted by adkinsjr View Post
    I'm not sure what rule your referring to. I would just decompose the function.
    yes both your method and my method output the values A=1 and B=(1/3). But plugging them back into the integral does not yield the correct answer, unless I'm missing a simplification step....

    My answer:
    \ln 3x + (1/3)\ln 1-x + c

    Book:

    <br />
(\frac {1}{3})\ln \frac {x}{(1-x)} + c<br />
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  6. #6
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    Recall the fact that

    \ln(a)- \ln(b) = \ln\left(\frac{a}{b}\right)
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  7. #7
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    Quote Originally Posted by pickslides View Post
    Recall the fact that

    \ln(a)- \ln(b) = \ln\left(\frac{a}{b}\right)
    I've considered that property, but each time I work out the problem, the natural logs come out as being added and not subtracted.
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  8. #8
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     <br />
\frac{1}{x} + \frac{1}{1-x} =\frac{1}{x} + \frac{1}{-(x+1)}= \frac{1}{x} - \frac{1}{x+1}<br />

    now apply the log law.
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  9. #9
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    Quote Originally Posted by skeeter View Post
    note that \frac{1}{3x(1-x)} = \frac{1}{3} \cdot \frac{1}{x(1-x)}

    and ...

    \frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x}<br />
    when the 1/3 is factored out I get:

    1 = A(1-x) + B(x)
     = x(B-A) + A

    then
     B - A = 0
    so
    B=A=1

    or am I doing it wrong?
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  10. #10
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    Quote Originally Posted by sgcb View Post
    I've considered that property, but each time I work out the problem, the natural logs come out as being added and not subtracted.
    I suggest skeeter's method as you don't need to solve partial fractions:

    <br />
\int \frac {1}{3x(1-x)} = \int \frac{1}{3} + \int \frac{1}{x} + \int \frac{1}{1-x}<br />

    ==============================================

    I've substituted the values of A and B you found into the expression \int\left(\frac{1}{3x}+\frac{1}{3-3x}\right)dx<br />

    Note that you integrate that second term you will need to divide by -3 because of the coefficient on the x

    \frac{1}{3}\ln(3x) - \frac{1}{3}\ln(3-3x) + C where k is a constant

    \frac{1}{3} \ln \left(\frac{3x}{3(1-x)}\right) +C = \frac{1}{3}\ln \left(\frac{x}{1-x}\right) + C
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  11. #11
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    \int \frac{1}{x} + \frac{1}{1-x} \, dx = \int \frac{1}{x} - \frac{-1}{1-x} \, dx = \ln|x| - \ln|1-x| + C = \ln\left|\frac{x}{1-x} \right| + C
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