# 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. $\displaystyle \frac{{3x^2 + x + 4}} {{\left( {x^2 + 1} \right)\left( {x^2 + 2} \right)}} = \frac{{ax + b}} {{x^2 + 1}} + \frac{{cx + d}} {{x^2 + 2}}$

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$

3. Originally Posted by Peritus
$\displaystyle \frac{{3x^2 + x + 4}} {{\left( {x^2 + 1} \right)\left( {x^2 + 2} \right)}} = \frac{{ax + b}} {{x^2 + 1}} + \frac{{cx + d}} {{x^2 + 2}}$

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.

4. $\displaystyle \begin{gathered} - \frac{1} {3}\int {\frac{{x + 2}} {{x^2 + x + 1}}dx = } - \frac{1} {3}\frac{1} {2}\int {\frac{{2\left( {x + 2} \right)}} {{x^2 + x + 1}}dx = } \hfill \\ - \frac{1} {3}\frac{1} {2}\int {\frac{{2x + 4}} {{x^2 + x + 1}}dx = } - \frac{1} {3}\frac{1} {2}\int {\frac{{2x + 1}} {{x^2 + x + 1}} + \frac{3} {{x^2 + x + 1}}dx} \hfill \\ \end{gathered}$

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