# Thread: Partial Fractions - Irreducible polynomials

1. ## 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?

2. $\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...

$
\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 = }
$

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

$
- \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 = }
$

$
= - \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
$

3. Originally Posted by Peritus
$\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...

$
\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 = }
$

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

$
- \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 = }
$

$
= - \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.

4. $
\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.