$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $ I'm trying to solve $$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $$
The value should be $\displaystyle \frac{\pi}{2}\ln(2)$. I tried different substitutions, but it doesn't work out. Changing the integration order seems not really nice as below:
$$\int_1^{\infty} \frac{2\sqrt{z+1}\arctan\left(\sqrt{\frac{z-1}{z+1}}\right)+\sqrt{z-1}(\ln(2z)-\ln(z-1))}{4z(z-1)(z+1)} dz$$
Has anyone a good start for me? Maybe Feynman? Thanks!

Suppose $\displaystyle\int\limits_0^x\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt$. It's well known that $\arctan z=\dfrac{\pi}{2}-\text{arccotan} z$, so our integral is $$\dfrac{\pi}{2}\int\limits_0^x\frac{dt}{(t^2+1)\sqrt{t^2+2}}\, dt-\int\limits_0^x\frac{\text{arccotan}\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt.$$
It's obvious that $\dfrac{1}{z}\text{arccotan}\dfrac{x}{z}=\dfrac{\pi}{2z}-\displaystyle\int_0^x\dfrac{dy}{y^2+z^2}$. Given that $z=\sqrt{2+t^2}$ and substituting the last equality in second integral, we get that our two integrals turns into $$\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+t^2+2)}=I.$$
Further, it's easy to see that $I=\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}-\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}$.
Since in the second integral everything is symmetric, $\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}=I$ (we can also make a replacement $t\to y$ and $y\to t$). Thus $$I=\dfrac{1}{2}\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}=\dfrac{1}{2}\left(\displaystyle\int\limits_0^x\dfrac{dt}{t^2+1}\right)^2=\dfrac{\arctan^2x}{2}.$$
It remains to calculate the integral $\dfrac{1}{2}\displaystyle\int\limits_0^\infty\dfrac{\arctan^2x}{x^2}dx=\displaystyle\int\limits_0^\infty\dfrac{\arctan x}{x(1+x^2)}dx$ (here we applying integrate by parts). Can you finish it yourself? Hint: you can consider the integral $\displaystyle\int\limits_0^\infty\dfrac{\arctan ax}{x(1+x^2)}dx$ and differentiate it by the parameter.
