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Thread: Partial Fractions - Irreducible polynomials

  1. #1
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    Partial Fractions - Irreducible polynomials

    Hi, any help would be much appreciated. I think I'm missing some vital theorem or technique because I have no idea how to solve the following.

    ∫(3x^2+x+4)/(x^4+3x^2+2)dx

    And

    ∫(-1/3x – 2/3) / (x^2+x+1)dx

    Is there a general technique to use? Any hints at all?

    Please and thank you.
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle \frac{{3x^2 + x + 4}}
    {{\left( {x^2 + 1} \right)\left( {x^2 + 2} \right)}} = \frac{{ax + b}}
    {{x^2 + 1}} + \frac{{cx + d}}
    {{x^2 + 2}}
    $

    can you continue...

    $\displaystyle
    \int {\frac{{ - \frac{1}
    {3}x - \frac{2}
    {3}}}
    {{x^2 + x + 1}}dx} = - \frac{1}
    {3}\int {\frac{{x + 2}}
    {{x^2 + x + 1}}dx} = - \frac{1}
    {6}\int {\frac{{2x + 1}}
    {{x^2 + x + 1}} + \frac{3}
    {{x^2 + x + 1}}dx = }
    $

    $\displaystyle
    = - \frac{1}
    {6}\ln \left| {x^2 + x + 1} \right| - \frac{1}
    {2}\int {\frac{1}
    {{x^2 + x + 1}}dx = }
    $


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

    $\displaystyle
    = - \frac{1}
    {6}\ln \left| {x^2 + x + 1} \right| - \frac{1}
    {2}\frac{4}
    {3}\sqrt {\frac{3}
    {4}} \arctan \left( {\frac{{x + {\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}}}
    {{\sqrt {\frac{3}
    {4}} }}} \right) + C
    $
    Last edited by Peritus; Feb 26th 2008 at 01:06 AM.
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  3. #3
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    Quote Originally Posted by Peritus View Post
    $\displaystyle \frac{{3x^2 + x + 4}}
    {{\left( {x^2 + 1} \right)\left( {x^2 + 2} \right)}} = \frac{{ax + b}}
    {{x^2 + 1}} + \frac{{cx + d}}
    {{x^2 + 2}}
    $

    can you continue...

    $\displaystyle
    \int {\frac{{ - \frac{1}
    {3}x - \frac{2}
    {3}}}
    {{x^2 + x + 1}}dx} = - \frac{1}
    {3}\int {\frac{{x + 2}}
    {{x^2 + x + 1}}dx} = - \frac{1}
    {6}\int {\frac{{2x + 1}}
    {{x^2 + x + 1}} + \frac{3}
    {{x^2 + x + 1}}dx = }
    $

    $\displaystyle
    = - \frac{1}
    {6}\ln \left| {x^2 + x + 1} \right| - \frac{1}
    {2}\int {\frac{1}
    {{x^2 + x + 1}}dx = }
    $


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

    $\displaystyle
    = - \frac{1}
    {6}\ln \left| {x^2 + x + 1} \right| - \frac{1}
    {2}\frac{4}
    {3}\sqrt {\frac{3}
    {4}} \arctan \left( {\frac{{x + {\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 2}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$2$}}}}
    {{\sqrt {\frac{3}
    {4}} }}} \right) + C
    $
    Thanks. I have a question though. How did you the bolded part? (2x + 1 that is). Is it when

    u = x^2 + x
    so du = 2x + 1? And if so, where did you get the 1/2 that was multiplied to 1/3 to make it 1/6.

    Thanks if you have time to answer.
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  4. #4
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    $\displaystyle
    \begin{gathered}
    - \frac{1}
    {3}\int {\frac{{x + 2}}
    {{x^2 + x + 1}}dx = } - \frac{1}
    {3}\frac{1}
    {2}\int {\frac{{2\left( {x + 2} \right)}}
    {{x^2 + x + 1}}dx = } \hfill \\
    - \frac{1}
    {3}\frac{1}
    {2}\int {\frac{{2x + 4}}
    {{x^2 + x + 1}}dx = } - \frac{1}
    {3}\frac{1}
    {2}\int {\frac{{2x + 1}}
    {{x^2 + x + 1}} + \frac{3}
    {{x^2 + x + 1}}dx} \hfill \\
    \end{gathered}

    $

    notice that the 0.5 and the 2 cancel each other so basically I don't change the integrand.
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