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Math Help - Integrating Partial Fractions

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    Integrating Partial Fractions

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    Quote Originally Posted by evansf View Post
    Evalulate ∫
    Hint:
    \frac{3x + 3}{x^3 - 1} = 3 \frac{x + 1}{(x - 1)(x^2 + x + 1)}

    So you are looking for partial fractions of the form
    \frac{x + 1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + c}{x^2 + x + 1}

    -Dan
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    Yes

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    Quote Originally Posted by evansf View Post
    Evalulate ∫
    Let's just focus on \frac{x+1}{(x-1)(x^2+x+1)}=\frac1{x^2+x+1}+\frac2{(x-1)(x^2+x+1)}.

    The first one is an arctangent. Let's see the second one:

    \frac{1}{{(x - 1)(x^2 + x + 1)}} = \frac{{x^2 + x + 1 - \left( {x^2 + x - 2} \right)}}{{3(x - 1)(x^2 + x + 1)}} = \frac{{x^2 + x + 1 - (x - 1)(x + 2)}}{{3(x - 1)(x^2 + x + 1)}}.

    This becomes to \frac{1}{3}\left( {\frac{1}{{x - 1}} - \frac{{x + 2}}{{x^2 + x + 1}}} \right).

    Now locate the second term:

    \frac{{x + 2}}{{x^2 + x + 1}} = \frac{1}{2}\left( {\frac{{2x + 4}}{{x^2 + x + 1}}} \right) = \frac{1}{2}\left( {\frac{{2x + 1}}{{x^2 + x + 1}} + \frac{3}{{x^2 + x + 1}}} \right).

    This is to avoid partial fractions. (It's just another way.)
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