Unable to crack $\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx=\frac{\pi^2}{30}$ I am unable to solve the integral
$$\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx=\frac{\pi^2}{30} $$
after a number of attempts, except with a few observations below.
1). Despite the appearance, I do not believe this integral is related to the Ahmed integral or any variations of known solution.
2). The integral is related to a more complex one posted here. However, the answer offered is non conventional, which I could not fully appreciate. I would prefer an integration approach.
3). Numerically, I could establish that
$$\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx
=2 \int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt5 x}{1+x^2}dx 
- \int_0^{\sqrt3} \frac{\cot^{-1}\sqrt5 x}{1+x^2}dx  $$
where the RHS can be worked out with an elaborate procedure.
So, I would like to see either a direct solution to the integral, or, at least, establish the integral relationship above analytically.

Perform the substitution $t=\frac{1}{\sqrt{2-x^2}}$ to check that,
$$
\tag{1}\int_{0}^{\frac{1}{\sqrt{3} } } \frac{\operatorname{arccot} \left ( \sqrt{2-x^2}  \right ) }{
1+x^2}\text{d}x=
\int_{\frac{\sqrt{2}}{2} }^{\sqrt{\frac{3}{5}}}
\frac{\arctan\left ( t \right ) }{\sqrt{2t^2-1}(3t^2-1) }\text{d}t.
$$
The second integral can be done by evaluating
$$
\tag{2}\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y
=\frac{\pi^2}{30}.
$$
To prove $(2)$, use the identity
$$
\int_{0}^{\infty} {e^{-t^2x^2}}\text{d}x
=\frac{\sqrt{\pi} }{2\sqrt{t} }.
$$
Then I can get(classical result $\int_{0}^{1} \frac{e^{-t^2x^2}}{1+t^2}\text{d}x
=\frac{\pi}{4}e^{t^2}\left ( 1-\operatorname{erf}(t)^2 \right )$ is used)
$$
\begin{align*}
&\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y\\
 =& \frac{2}{\sqrt{\pi} } \int_{0}^{1} \int_{0}^{1} 
\frac{1}{(1+x^2)(1+y^2)} \left (\int_{0}^{\infty}e^{-t^2(3+x^2+y^2)}\text{d}t\right ) \text{d}x\text{d}y\\
\overset{{\scriptsize\text{Fubini}}}{=}&
\frac{2}{\sqrt{\pi} } \int_{0}^{\infty}e^{-3t^2}\left (  \int_{0}^{1} \frac{e^{-t^2x^2}}{1+x^2}\text{d} x\right )^2\text{d}t\\
=&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16}  \int_{0}^{\infty}
e^{-x^2}\left(1-\operatorname{erf}(x)^2\right)^2\text{d}x\\
=&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16}  \int_{0}^{\infty}
e^{-x^2}\left(1-2\operatorname{erf}(x)^2+\operatorname{erf}(x)^4\right)\text{d}x\\
=&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16}\left(\int_{0}^{\infty}e^{-x^2}\text{d}x
-2\int_{0}^{\infty}e^{-x^2}\operatorname{erf}(x)^2\text{d}x
+\int_{0}^{\infty}e^{-x^2}\operatorname{erf}(x)^4\text{d}x\right)\\
=&\frac{\pi^2}{30}.
\end{align*}
$$
The last equality follows from their primitives
$$
\frac{\mathrm{d} }{\mathrm{d} x} 
\left ( \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)^{n+1} \right )
=(n+1)e^{-t^2}\operatorname{erf}(x)^{n}.
$$
And
$$
\begin{align*}
&\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y\\
=&\int_{0}^{1} \frac{\arctan\left ( \sqrt{\frac{2+x^2}{4+x^2} }  \right )  }{
\left ( 1+x^2 \right )\sqrt{2+x^2}  }\text{d}x 
\\
\overset{{\scriptsize t=\sqrt{\frac{2+x^2}{4+x^2} }}}{=}&\int_{\frac{\sqrt{2}}{2} }^{\sqrt{\frac{3}{5}}}
\frac{\arctan\left ( t \right ) }{\sqrt{2t^2-1}(3t^2-1) }\text{d}t\\
=&\int_{0}^{\frac{1}{\sqrt{3} } } \frac{\operatorname{arccot} \left ( \sqrt{2-x^2}  \right ) }{
1+x^2}\text{d}x=\frac{\pi^2}{30}.
\end{align*}
$$
