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Math Help - Integration by Partial Fraction Decomposition

  1. #1
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    Integration by Partial Fraction Decomposition

    I can find the function that needs to be integrated and get the same function as in the answers but I canít seem to find a way to integrate the function. I have tried many times and failed, so could someone help me through or show me how its done as I am out of ideas.

    Thanks, Daddy_Long_Legs
    Attached Thumbnails Attached Thumbnails Integration by Partial Fraction Decomposition-question-1.-h-working.jpg  
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  2. #2
    MHF Contributor red_dog's Avatar
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    \frac{1}{x^3-1}=\frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}

    Now you have to find A, B, C.
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  3. #3
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    Hi red_dog

    I have said I can already find the function that needs to be integrated but I canít seem to find a way to integrate the function. That is I have found the values of A, B and C. Iím just not sure how to integrate it when I have this function. Look at the attachment please to see what I mean.
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  4. #4
    MHF Contributor red_dog's Avatar
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    \frac{1}{3}\int\frac{1}{x-1}dx-\frac{1}{3}\int\frac{x+2}{x^2+x+1}dx=

    =\frac{1}{3}\ln|x-1|-\frac{1}{6}\int\frac{2x+4}{x^2+x+1}dx=

    =\frac{1}{3}\ln|x-1|-\frac{1}{6}\left(\int\frac{2x+1}{x^2+x+1}dx+3\int\  frac{dx}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}  \right)=

    =\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln(x^2+x+1)-\frac{1}{2}\cdot\frac{2}{\sqrt{3}}\arctan\left(\fr  ac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C=

    =\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln(x^2+x+1)-\frac{1}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}+C
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  5. #5
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    Thanks red_dog

    I tried doing something similar to by completing the square but this was where I ran into the trouble. I should be able to the other questions that are similar to this so thanks for the help.
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