Integrating $\int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$ I found the following integral and wanted to know if there is a nice closed form solution in terms of elementary or some special functions (Polylogarithm, Clausen, etc).
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$$
I know that the integral converges numerically to $\approx 0.403926$
Here is my try using integration by parts:
Let
$$ du = \frac{\arctan(x)}{x^2} \Longrightarrow u = -\frac{1}{2}\ln(1+x^2) + \ln(x) - \frac{\arctan(x)}{x} $$
$$ v = \arctan(x^2) \Longrightarrow dv = \frac{2x}{x^4+1}$$
Hence
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx \stackrel{IBP}{=} -\frac{\pi^2}{16} - \frac{1}{8}\pi \ln(2)  -2\underbrace{\int_{0}^{1} \frac{x\ln(x)}{x^4+1} dx}_{I_{1}} + 2\underbrace{\int_{0}^{1}\frac{\arctan(x)}{x^4+1}dx}_{I_{2}} + \underbrace{\int_{0}^{1} \frac{x\ln(1+x^2)}{x^4+1}dx}_{I_{3}} $$
Can be proven that
$$I_{1} = -\frac{C}{4}$$
where $C$ is the Catalan constant and
$$I_{3} = \frac{1}{16} \pi \ln(2) $$
but I'm stuck with $I_{2}$
Another way could be:
Define
$$\Psi(a) =  \int_{0}^{1} \frac{\arctan(ax)\arctan(x^2)}{x^2} dx$$
Hence
$$\Psi'(a) = \int_{0}^{1} \frac{\arctan(x^2)}{x(a^2x^2+1)}dx = \frac{1}{2}\int_{0}^{1} \frac{\arctan(w)}{w(a^2w+1)}dx$$
Using integration by parts, we have:
$$du = \frac{\arctan(w)}{1+a^2w} \Longrightarrow u= \frac{1}{1+a^4}\ln\left( \frac{1+a^2w}{\sqrt{1+w^2}} \right) - \frac{a^2-w}{(1+a^4)(1+a^2w)} \arctan(w)  $$
$$ v = \frac{1}{w} \Longrightarrow dv = \ln(w) $$
However, this path seems even more rugged that the other.
One last hint could be the following integral:
$$\int_{0}^{1} \frac{\arctan(x) \arctan(x^3)}{x} dx = \frac{7}{72}\zeta(3) + \frac{\pi}{3}C - \frac{5\pi}{12}\operatorname{Cl}_{2}\left(\frac{2\pi}{3} \right)$$
where $\operatorname{Cl}_{2}$ is the Clausen function of order 2. However, I do not know the proof of this result either.

By continuing the OP's work and using Cornel's closed-form of $\displaystyle \int_0^1\frac{\arctan(x)}{1+x^4}\textrm{d}x$, we get that

$$\int_0^1\frac{\arctan(x)\arctan(x^2)}{x^2}\textrm{d}x$$
$$=G+\frac{11}{192}\sqrt{2}\pi^2-\frac{1}{16}\log(2)\pi-\frac{3}{16}\sqrt{2}\log(\sqrt{2}-1)\pi-\frac{1}{4}\sqrt{2}\operatorname{Li}_2(1-\sqrt{2})-\frac{1}{32}\psi^{(1)}\left(\frac{1}{8}\right).$$

