Closed form of an integral

Question

Is there a closed form of $$\int\limits_0^1\frac{\arctan(\sqrt{x^2+2})}{(1+x^2)\cdot(\sqrt{x^2+2})}dx \quad \text{?}$$

I just know that $\int\limits_0^1\frac{\arctan(\sqrt{x^2+2})}{(1+x^2)\cdot(\sqrt{x^2+2})}dx = 0.514042...$

Answer 1

This is called the Ahmed integral.

$$ \int_{0}^{1} \frac{\tan^{-1}\sqrt{x^2 + 2}}{(x^2 + 1)\sqrt{x^2 + 2}} \, dx = \frac{5\pi^2}{96} . $$

You can see a solution here.