How to evaluate $\int { \frac { dx }{ { 2x }^{ 4 }+{ 2x }^{ 2 }+1 } }$? How to integrate something like 
$\int { \frac { dx }{ { 2x }^{ 4 }+{ 2x }^{ 2 }+1 }  }$
I know the method of partial fractions.
But what if the denominator is not factorizeable like the one in the question?
 A: A partial partial fraction approach:
$$\frac{1}{ax^4+bx^2+c} = \frac{1}{a}\left(\frac{A_1}{x^2-r_1} + \frac{A_2}{x^2-r_2}\right)$$
where $r_1,r_2$ are the (possibly complex) roots of $au^2+bu+c=0$.
(You'll have to take care if $r_1=r_2$.)
Solve for $A_1,A_2$, and then use techniques know for integrating $\int \frac{dx}{x^2-r_i}$.
Ultimately, you need to know how to compute: $\log(x-(u+vi))$. Here, we get:
$$\log(x-(u+vi)) = \log(\sqrt{(x-u)^2+v^2})+i\theta$$ there $\tan\theta = \frac{v}{x-u}$. The complex parts will cancel out, so $\theta$ is irrelevant.
A: First pull out any constant coefficient of $x^4$ to get a monic polynomial.
In $$ x^4 + B x^2 + C  $$ with $$ B^2 - 4 C < 0,  $$ so that $C > 0,$ we get
$$ x^4 + B x^2 + C = (x^2 + \lambda x + \sqrt C)  (x^2 - \lambda x + \sqrt C),  $$
where
$$ \lambda = \sqrt {2 \sqrt C - B}  $$
is also real.
The original problem was edited to $2 x^4 + 2 x^2 + 1 = 2 \left(x^4 + x^2 + \frac{1}{2} \right);$
$$ x^4 + x^2 + \frac{1}{2} = \left(x^2 + \sqrt{\sqrt 2 - 1} \; x +  \frac{1}{\sqrt 2} \right)  \left(x^2 -  \sqrt{\sqrt 2 - 1} \; x + \frac{1}{\sqrt 2} \right)   $$
