# Integral ∫10arctan2x√1−x2dx\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx

Is it possible to evaluate this integral in a closed form?

It also can be represented as

Okay, finally I was able to prove it.

Step 0. Observations. In view of the following identity

Vladimir’s result suggests that there may exists a general formula connecting

and the Legendre chi function $\chi_{2}$. Indeed, inspired by Vladimir’s result, I conjectured that

I succeeded in proving this identity, so I post a solution here.

Step 1. Proof of the identity $\text{(1)}$. It is easy to check that the following identity holds:

So it follows that

For the convenience of notation, we put

Then it is easy to check that $\mathrm{arsinh}(1/r) = - \log \alpha$ and likewise for $s$ and $\beta$. Thus with the substitution $x \mapsto \sinh x$ and $y \mapsto \sinh y$, we have

Applying the substitution $e^{-x} \mapsto x$ and $e^{-y} \mapsto y$, it follows that

as desired, proving the identity $\text{(1)}$.

EDIT. I found a much simpler and intuitive proof of $\text{(1)}$. We first observe that $\text{(1)}$ is equivalent to the following identity

Now we first observe that from the addition formula for the hyperbolic tangent, we obtain the following formula

which holds for sufficiently small $x, y$. Thus

We readily check this holds for any $|r| < 1$. Therefore