$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$ Without Feynman Integration

Question

How do I find $$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$$
without Feynman integration? I saw this video, which gives $$\int_{0}^{1}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\frac{5\pi^2}{96}$$
Via Feynman integration, but I would like to know another method.

Answer 1

Notice that $\frac{\arctan x}{x} = \int_{0}^{1} \frac{dy}{1+x^2y^2}$. So

\begin{align*} \int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx &= \int_{-\infty}^{\infty} \int_{0}^{1} \frac{1}{(x^2+1)(1+(x^2+2)y^2)} \, dydx. \end{align*}

Notice that

$$ \frac{1}{(x^2+1)(1+(x^2+2)y^2)} = \frac{1}{(x^2+1)(y^2+1)} - \frac{y^2}{(y^2+1)(1+(x^2+2)y^2)}. $$

So interchanging the order of integration,

\begin{align*} \int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx &= \int_{0}^{1} \left( \frac{\pi}{y^2+1} - \frac{\pi y}{(y^2+1)\sqrt{2y^2+1}} \right) \, dy \\ &= \pi \left[ \arctan(y) - \arctan\sqrt{2y^2+1} \right]_{0}^{1} \\ &= \frac{\pi^2}{6}. \end{align*}