How to calculate the singularities of $x^4 + x^2 + 1$?

Question

Here is the integration I want to calculate (complex analysis):

$$\int_{-\infty}^{\infty} \frac{\cos x}{x^4 + x^2 +1}$$

But I do not know how to factorize the following equation $x^4 + x^2 + 1$ to get the singularities. I first let $y = x^2$ and I used the quadratic formula to get $y = \frac{-1 \pm \sqrt{3}i}{2}$ but then what should I do to get $x$? Could someone help me please?

EDIT:

I think my professor did it using that any complex number can be written as $r e^{i \theta}$ but I do not know how he got the $\theta$, could anyone show me the solution by this method please?

Answer 1

Method I

$x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ $=(x-a_1)(x-a_2)(x-a_3)(x-a_4)$

where $a_1,a_2,a_3,a_4$ are roots of those two quadratic equations.

Method II

Start from your $x^2=y= \frac{-1 \pm \sqrt{3}i}{2}=cos\pm\frac{2\pi}{3}+isin\pm\frac{2\pi}{3}=cos\pm\frac{4\pi}{3}+isin\pm\frac{4\pi}{3}$

Then you can get 4 $x$'s like

$x=cos\pm\frac{\pi}{3}+isin\pm\frac{\pi}{3}, cos\pm\frac{2\pi}{3}+isin\pm\frac{2\pi}{3}$

Answer 2

You can multuply $x^4+x^2+1$ with $x^2 - 1$ and solve equation $x^6=1$ then remember that extrude root $x = \pm 1$.

Answer 3

You can compute the square roots of your $y_\pm:=\frac{-1\pm i\sqrt3}2$ directly, by the classical method indicated in the proposed duplicate. E.g. for $y_+$: let $z=a+ib$ with $a,b\in\mathbb R$. $$\begin{align}z^2=\frac{-1+i\sqrt3}2&\Leftrightarrow a^2-b^2=-\frac12,\;a^2+b^2=1,\;2ab>0\\ &\Leftrightarrow a^2=\frac14,\;b^2=\frac34,\;ab>0\\ &\Leftrightarrow z=\pm\frac{1+i\sqrt3}2.\end{align}$$