Show that : $\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$ I found it with help of Wolfram alpha .
$$\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$$
This integral admits an antiderivative (I can add it if so) so I can conclude that this result is true .I would like to see a proof with the use of complex integration or a 'real' proof . For that I think we can put $y=\frac{2}{x^2+1}$ and use integration by parts (perhaps?).For the proof with complex integration(if it exists I'm new for it )  can you detail the solution ?
Thanks!

$$I=\int_{0}^{\infty} \tan^{-1}\frac{2}{x^2+1} dx$$
Let us do it by parts taking 1 as second function:
$$I=x\tan^{-1}\frac{2}{1+x^2}|_{0}^{\infty}+4\int_{0}^{\infty}\frac{x^2}{x^4+2x^2+5}dx$$
$$\implies I=4\int_{0}^{\infty} \frac{dx}{x^2+5/x^2+2}$$ $$=2\int_{0}^{\infty} \left(\frac{(1+\sqrt{5}/x^2)}{(x-\sqrt{5}/x)^2+2+2\sqrt{5}}+\frac{(1-\sqrt{5}/x^2)}{(x+\sqrt{5}/x)^2+2-2\sqrt{5}}\right)dx$$
Let $x-\sqrt{5}/x=u$ and $x+\sqrt{5}/x=v$, then the limits in the second integral being identical it vanishes, we get
$$I=2\int_{-\infty}^{\infty} \frac{du}{u^2+2+2\sqrt{5}}=\frac{2}{\sqrt{2+2\sqrt{5}}} \tan^{-1}\frac{u}{\sqrt{2+2\sqrt{5}}}|_{-\infty}^{\infty}=\pi\sqrt{\frac{2}{1+\sqrt{5}}}=\pi \phi^{-1/2}.$$
