integration (x^2+1)/(x^4+x^2+1)

$\begin{aligned} I & = \int_0^\infty \frac {x^2+1}{x^4+x^2+1} dx \\ & = \int_0^\infty \frac {x^2+1}{(x^2+x+1)(x^2-x+1)} dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{x^2+x+1} + \frac 1{x^2-x+1} \right) dx \\ & = \frac 23 \left(\int_0^\infty \frac {dx}{\left(\frac {2x+1}{\sqrt 3}\right)^2+1} + \int_0^\infty \frac {dx}{\left(\frac {2x-1}{\sqrt 3}\right)^2+1} dx \right) & \small \blue{\text{Let }\tan \theta \frac {2x \pm 1}{\sqrt 3} \implies \sec^2 \theta = \frac 2{\sqrt 3} dx} \\ & = \frac 1{\sqrt 3} \left( \int_0^\frac \pi 2 d\theta + \int_0^\frac \pi 2 d\theta \right) \\ & = \frac \pi {\sqrt 3} \approx \boxed{1.81} \end{aligned}$

Factorising the denominator we get the indefinite integral as

$\dfrac {1}{2}\left (\displaystyle \int \frac{dx}{x^2+x+1}+\displaystyle \int \frac{dx}{x^2-x+1}\right )$

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

Hence the value of the definite integral is $\dfrac {π}{\sqrt 3}\approx \boxed {1.813799}$ .

Here is a bit different from othere two.

$\int _0^\infty \dfrac{x^2+1}{x^4+x^2+1}{d}x$

$=\int_0^\infty \dfrac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}{d}x$

$=\int_0^\infty \dfrac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+3}{d}x$

$= \int_{-\infty}^{\infty} \dfrac{1}{t^2+3}{d}t$ {Put $(x-\frac{1}{x}) = t$ }

$=\dfrac{1}{\sqrt{3}}\left.\arctan\frac{x}{\sqrt(3)}\right|_-\infty^\infty$

$=\dfrac{\pi}{3^1/2}$ $\blacksquare$

ps.. I had the heck of a time typing the latex myself.

@Admins please enable the toggle Latex feature again.