# How do you express 1/(x^2+x+1)  in partial fractions?

Mar 24, 2016

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

#### Explanation:

This expression can only be split down further if we use Complex coefficients...

${x}^{2} + x + 1$

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

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

$= \left(x + \frac{1}{2} - \frac{\sqrt{3}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

So:

$\frac{1}{{x}^{2} + x + 1} = \frac{A}{x + \frac{1}{2} - \frac{\sqrt{3}}{2} i} + \frac{B}{x + \frac{1}{2} + \frac{\sqrt{3}}{2} i}$

$= \frac{A \left(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i\right) + B \left(x + \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)}{{x}^{2} + x + 1}$

$= \frac{\left(A + B\right) x + \left(A \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) + B \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)\right)}{{x}^{2} + x + 1}$

Equating coefficients, we find:

$\left\{\begin{matrix}A + B = 0 \\ A \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) + B \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) = 1\end{matrix}\right.$

From the first equation we find $B = - A$.

Substitute this in the second equation to get:

$A \sqrt{3} i = 1$

Hence:

$\left\{\begin{matrix}A = \frac{1}{\sqrt{3} i} = - \frac{\sqrt{3}}{3} i \\ B = \frac{\sqrt{3}}{3} i\end{matrix}\right.$

So:

$\frac{1}{{x}^{2} + x + 1} = \frac{B}{x + \frac{1}{2} + \frac{\sqrt{3}}{2} i} + \frac{A}{x + \frac{1}{2} - \frac{\sqrt{3}}{2} i}$

$= \frac{\sqrt{3} i}{3 \left(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)} - \frac{\sqrt{3} i}{3 \left(x + \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)}$