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Math Help - Integration partial fractions

  1. #1
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    Integration partial fractions

    Hi all,
    IS this correct?
    thanks
    John

    [math}\int \frac{x.dx}{x^2+4x+13}[/tex]
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  2. #2
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    Re: Integration partial fractions

    Quote Originally Posted by celtic1234 View Post
    Hi all,
    IS this correct?
    thanks
    John

    \displaystyle \int \frac{10}{(2x-6)(x+5)}
    You've only stated the question? Partial fractions is a good method here but first some algebra:

    \int \dfrac{10}{(2x-6)(x+5)} = \int \dfrac{10}{2(x-3)(x+5)} = 5 \int \dfrac{1}{(x-3)(x+5)}


    \dfrac{1}{(x-3)(x+5)} = \dfrac{A}{x-3} + \dfrac{B}{x+5} \Leftrightarrow 1 = A(x+5) + B(x-3)

    Can you continue?


    edit: That other integral is not an easy one to work out. Are you familiar with trig integrals?
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  3. #3
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    Re: Integration partial fractions

    sorry i am busy in the background trying to get laTex to work-i'll try and sort it ASAP
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    Re: Integration partial fractions

    Quote Originally Posted by celtic1234 View Post
    sorry i am busy in the background trying to get laTex to work-i'll try and sort it ASAP
    Your latex is good. You just need to change the tag from [tex]...[/tex] to [tex]...[/tex]
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  5. #5
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    Re: Integration partial fractions

    I am solving the partial fraction to give A= 5/4 and B=-1/6.

    The integral solves as:

     \frac{1}{4}5ln(2x-6) - \frac{1}{6}ln(x+1) +c

    Can you tell me if i am correct?
    thanks
    John
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    Re: Integration partial fractions

    Quote Originally Posted by celtic1234 View Post
    I am solving the partial fraction to give A= 5/4 and B=-1/6.

    The integral solves as:

     \frac{1}{4}5ln(2x-6) - \frac{1}{6}ln(x+1) +c

    Can you tell me if i am correct?
    thanks
    John
    I get B = -1/8

    Sub in x=-5: 1 = B(-5-3) = -8B \Leftrightarrow B = -\dfrac{1}{8}

    Sub in x=3: 1 = 8A \Leftrightarrow A = \dfrac{1}{8}

    Hence 5 \int \dfrac{1}{(x-3)(x+5)} = 5 \left(\int \dfrac{1}{8(x-3)} - \int \dfrac{1}{8(x+5)}\right) = \dfrac{5}{8} \left(\int \dfrac{1}{x-3} - \int \dfrac{1}{x+5}\right)

    =\dfrac{5}{8} \left( \ln(x-3) - \ln(x+5) + C_1\right) = \dfrac{5}{8} \ln \left(\dfrac{x-3}{x+5}\right) + C \text{  where  } C = \dfrac{5}{8}C_1


    Perhaps it would be easier to make a new thread since there seem to be two questions on the go here! Alternatively I can delete my posts regarding the original integral
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  7. #7
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    Re: Integration partial fractions

    ok thanks
    i'll do a new thread
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