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Math Help - 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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    \frac{{3x^2  + x + 4}}<br />
{{\left( {x^2  + 1} \right)\left( {x^2  + 2} \right)}} = \frac{{ax + b}}<br />
{{x^2  + 1}} + \frac{{cx + d}}<br />
{{x^2  + 2}}<br />

    can you continue...

    <br />
\int {\frac{{ - \frac{1}<br />
{3}x - \frac{2}<br />
{3}}}<br />
{{x^2  + x + 1}}dx}  =  - \frac{1}<br />
{3}\int {\frac{{x + 2}}<br />
{{x^2  + x + 1}}dx}  =  - \frac{1}<br />
{6}\int {\frac{{2x + 1}}<br />
{{x^2  + x + 1}} + \frac{3}<br />
{{x^2  + x + 1}}dx = } <br />

    <br />
 =  - \frac{1}<br />
{6}\ln \left| {x^2  + x + 1} \right| - \frac{1}<br />
{2}\int {\frac{1}<br />
{{x^2  + x + 1}}dx = } <br />


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

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

    can you continue...

    <br />
\int {\frac{{ - \frac{1}<br />
{3}x - \frac{2}<br />
{3}}}<br />
{{x^2  + x + 1}}dx}  =  - \frac{1}<br />
{3}\int {\frac{{x + 2}}<br />
{{x^2  + x + 1}}dx}  =  - \frac{1}<br />
{6}\int {\frac{{<b>2x + 1</b>}}<br />
{{x^2  + x + 1}} + \frac{3}<br />
{{x^2  + x + 1}}dx = } <br />

    <br />
 =  - \frac{1}<br />
{6}\ln \left| {x^2  + x + 1} \right| - \frac{1}<br />
{2}\int {\frac{1}<br />
{{x^2  + x + 1}}dx = } <br />


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

    <br />
 =  - \frac{1}<br />
{6}\ln \left| {x^2  + x + 1} \right| - \frac{1}<br />
{2}\frac{4}<br />
{3}\sqrt {\frac{3}<br />
{4}} \arctan \left( {\frac{{x + {\raise0.7ex\hbox{$1$} \!\mathord{\left/<br />
 {\vphantom {1 2}}\right.\kern-\nulldelimiterspace}<br />
\!\lower0.7ex\hbox{$2$}}}}<br />
{{\sqrt {\frac{3}<br />
{4}} }}} \right) + C<br />
    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
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    Posts
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    <br />
\begin{gathered}<br />
   - \frac{1}<br />
{3}\int {\frac{{x + 2}}<br />
{{x^2  + x + 1}}dx = }  - \frac{1}<br />
{3}\frac{1}<br />
{2}\int {\frac{{2\left( {x + 2} \right)}}<br />
{{x^2  + x + 1}}dx = }  \hfill \\<br />
   - \frac{1}<br />
{3}\frac{1}<br />
{2}\int {\frac{{2x + 4}}<br />
{{x^2  + x + 1}}dx = }  - \frac{1}<br />
{3}\frac{1}<br />
{2}\int {\frac{{2x + 1}}<br />
{{x^2  + x + 1}} + \frac{3}<br />
{{x^2  + x + 1}}dx}  \hfill \\ <br />
\end{gathered} <br /> <br />

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