How to evaluate $\int_{0}^{\infty} \frac{\arctan x}{x^2+x+1}dx$

Question

$$\int_{0}^{\infty} \frac{\arctan x}{x^2+x+1}dx$$ 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

Answer 1

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 $$\int_a^bf(x)\ dx=-\int_b^af(x)\ dx$$

Answer 2

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, $$f(\tfrac12)=\frac{\pi^2}{6\sqrt3}.$$