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Math Help - Another partial fraction

  1. #1
    Junior Member symstar's Avatar
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    Another partial fraction

    \int\frac{1}{x^3-1}dx

    Fraction work:
    \frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}
    1=(A+B)x^2+(A-B+C)x+(A-C)
    A=\tfrac{1}{3}; B=\tfrac{-1}{3}; C=\tfrac{2}{3}

    Back to the integral:
    \frac{1}{3}\int\frac{dx}{x-1}+\int\frac{\tfrac{-1}{3}x+\tfrac{2}{3}}{x^2+x+1}dx
    \tfrac{1}{3}\ln{\left|x-1\right|}-\frac{1}{3}\int\frac{xdx}{x^2+x+1}+\frac{2}{3}\int  \frac{dx}{x^2+x+1}

    I considered u=x^2+x+1 \text{ ,but that gives } du=2x+1dx which doesn't seem to help the situation much. Should completing the square be a consideration?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by symstar View Post
    \int\frac{1}{x^3-1}dx

    Fraction work:
    \frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}
    1=(A+B)x^2+(A-B+C)x+(A-C)
    A=\tfrac{1}{3}; B=\tfrac{-1}{3}; C=\tfrac{2}{3}

    Back to the integral:
    \frac{1}{3}\int\frac{dx}{x-1}+\int\frac{\tfrac{-1}{3}x+\tfrac{2}{3}}{x^2+x+1}dx
    \tfrac{1}{3}\ln{\left|x-1\right|}-\frac{1}{3}\int\frac{xdx}{x^2+x+1}+\frac{2}{3}\int  \frac{dx}{x^2+x+1}

    I considered u=x^2+x+1 \text{ ,but that gives } du=2x+1dx which doesn't seem to help the situation much. Should completing the square be a consideration?
    yes, complete the square for the last two integrals
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  3. #3
    Junior Member symstar's Avatar
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    Take just the integral: -\frac{1}{3}\int\frac{xdx}{x^2+x+1}

    Completing the square:

    -\frac{1}{3}\int\frac{xdx}{(x+\tfrac{1}{2})^2+\tfra  c{3}{4}}
    u=x+\tfrac{1}{2} \text{ so } du=dx \text{ and }x=u-\tfrac{1}{2}
    -\frac{1}{3}\int\frac{u-\tfrac{1}{2}du}{u^2+\tfrac{3}{4}}

    Is this correct? This seems to just be leading me in to more of a mess.
    Last edited by symstar; September 14th 2008 at 06:39 PM. Reason: added stuff
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by symstar View Post
    Take just the integral: -\frac{1}{3}\int\frac{xdx}{x^2+x+1}

    Completing the square:

    -\frac{1}{3}\int\frac{xdx}{(x+\tfrac{1}{2})^2+\tfra  c{3}{4}}
    u=x+\tfrac{1}{2} \text{ so } du=dx \text{ and }x=u-\tfrac{1}{2}
    -\frac{1}{3}\int\frac{u-\tfrac{1}{2}du}{u^2+\tfrac{3}{4}}

    Is this correct? This seems to just be leading me in to more of a mess.
    yes, you are good so far. now factor out the 3/4 from the denominator and split the integral.
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  5. #5
    Member javax's Avatar
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    Or with -\frac{1}{3}\int{\frac{x dx}{x^2+x+1 }} you could have go
     <br />
-\frac{1}{3\cdot \color{red}{2}}\int{\frac{\color{red}{2}\color{bla  ck}{x+1}}{x^2+x+1}}dx-\int{\frac{dx}{x^2+x+1}}<br />
    I believe now is easier
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  6. #6
    Junior Member symstar's Avatar
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    Carrying on...
    -\frac{1}{3}\left(\int\frac{u}{u^2+\tfrac{3}{4}}du-\frac{1}{2}\int\frac{du}{u^2+\tfrac{3}{4}}\right)
    -\frac{1}{3}\left(\frac{1}{2}\ln{(u^2+\tfrac{3}{4})  }-\frac{1}{2}(\frac{2}{\sqrt{3}})\tan^{-1}(\tfrac{2}{\sqrt{3}}(u^2+\tfrac{3}{4})\right)
    -\tfrac{1}{6}\ln{(x^2+x+1)}-\tfrac{1}{\sqrt{3}}\tan^{-1}(\tfrac{2}{\sqrt{3}}(x^2+x+1)

    If I did indeed do this correctly, then I still have the \frac{2}{3}\int\frac{dx}{x^2+x+1} from earlier which I still need to integrate. According to the book, I'll get the incorrect answer
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