How to evaluate $\int_{0}^{\infty} \frac{\arctan x}{x^2+x+1}dx$ $$\int_{0}^{\infty} \frac{\arctan x}{x^2+x+1}dx$$
The only idea I have is to formulate the denominator, get the derivative form of $\arctan$ and then perform the segmentation, but this doesn't seem to work.
How to use calculus to calculate this integral?
Any help will be appreciated
 A: Indeed, we may evaluate
$$f(a)=\int_0^\infty\frac{\arctan x}{x^2+2ax+1}dx\qquad |a|<1.$$
Using $x=1/t$,
$$f(a)=\int_0^\infty \frac{\pi/2-\arctan t}{\frac1{t^2}+\frac{2a}{t}+1}\frac{dt}{t^2}=\frac\pi2\int_0^\infty\frac{dt}{t^2+2a+1}-f(a).$$
Thus 
$$f(a)=\frac\pi4\int_0^\infty\frac{dx}{x^2+2ax+1}.$$
We complete the square in the denominator,
$$f(a)=\frac\pi4\int_0^\infty\frac{dx}{(x+a)^2+1-a^2}$$
and note that $|a|<1$ ensures that $1-a^2>0$ so that we may set $x+a=\sqrt{1-a^2}\tan t$,
$$f(a)=\frac{\pi}{4\sqrt{1-a^2}}\int_{\phi(a)}^{\pi/2}\frac{\sec^2 t\ dt}{1+\tan^2 t}=\frac\pi{4\sqrt{1-a^2}}\left(\frac\pi2-\phi(a)\right)$$
where 
$$\phi(a)=\arctan\frac{a}{\sqrt{1-a^2}}.$$
For the case $a=1/2$, which is your integral,
$$f(\tfrac12)=\frac{\pi^2}{6\sqrt3}.$$
A: Set $\dfrac1x=y$
$$I=\int_{0}^{\infty} \frac{\arctan x}{x^2+x+1}dx=\int_\infty^0\dfrac{\arctan\dfrac1y}{\dfrac1{y^2}+\dfrac1y+1}\left(-\dfrac1{y^2}\right)=\int_0^\infty\dfrac{\dfrac\pi2}{y^2+y+1}-I$$
using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?  and $\arctan x+\text{arccot}x=\dfrac\pi2$
and $$\int_a^bf(x)\ dx=-\int_b^af(x)\ dx$$
