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Math Help - Intergration of fraction

  1. #1
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    Intergration of fraction

    Intergrate -x/(1-x^2)dx, i tried partial fractions but i didn't get a valid answer,

    Thanks for any help
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Intergration of fraction

    I would try a u-substitution:

    u=1-x^2\,\therefore\,du=-2x\,dx
    Thanks from Saxby
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  3. #3
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    Re: Intergration of fraction

    Yep i've got the answer, -(1/2)ln(1-x^2). Thanks for the suggestion
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Intergration of fraction

    You don't want the negative sign out front...

    \int\frac{-x}{1-x^2}\,dx=\frac{1}{2}\int\frac{-2x\,dx}{1-x^2}=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|  +C=\frac{1}{2}\ln|1-x^2|+C
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  5. #5
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    Re: Intergration of fraction

    Hello, Saxby!

    \int\frac{ -x}{1-x^2}\,dx

    I tried partial fractions but i didn't get a valid answer. .Must have made errors.

    We have: . \int\frac{-x}{1-x^2}\,dx \:=\:\int\frac{x}{x^2-1}\,dx \:=\:\int\frac{x}{(x-1)(x+1)}\,dx


    \frac{x}{(x-1)(x+1)} \:=\:\frac{A}{x-1} + \frac{B}{x+1} \quad\Rightarrow\quad x \:=\:A(x+1) + B(x-1)

    Let x = 1\!:\;1 \:=\:A(2) + B(0) \quad\Rightarrow\quad A = \tfrac{1}{2}

    Let x = \text{-}1\!:\;\text{-}1 \:=\:A(0) + B(\text{-}2) \quad\Rightarrow\quad B = \tfrac{1}{2}


    We have: . \int\left(\frac{\frac{1}{2}}{x-1} + \frac{\frac{1}{2}}{x+1}\right)\,dx \;=\; \tfrac{1}{2}\int\left(\frac{1}{x-1} + \frac{1}{x+1}\right)\,dx

    . . . . . . . =\;\tfrac{1}{2}\big(\ln|x-1| + \ln|x+1|\big)+C \;=\;\tfrac{1}{2}\ln|(x-1)(x+1)| + C

    . . . . . . . =\;\tfrac{1}{2}\ln\left|x^2-1\right| + C
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  6. #6
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    Re: Intergration of fraction

    Hi guys , happy to see you here

    partial fractions can be made easier ,if we played a little trick

    \frac{1}{2} \int \frac{2x}{x^2-1}\, dx =\frac{1}{2} \int \frac{1+x+x-1}{x^2-1}\, dx=\frac{1}{2} \int \frac{x+1}{x^2-1}+\frac{x-1}{x^2-1}\, dx

    =\frac{1}{2} \int \frac{1}{x-1}+\frac{1}{x+1}\, dx
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: Intergration of fraction

    Hey there! Glad to see a familiar name!

    We could avoid the negative sign issue altogether by writing:

    \int\frac{-x}{1-x^2}\,dx=\int\frac{x}{x^2-1}\,dx=\frac{1}{2}\int\frac{2x\,dx}{x^2-1}=\frac{1}{2}\int\frac{du}{u}=

    \frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|x^2-1|+C
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