This integral appears very similar to $\int\frac{\arctan x}{1+x^2}\,\mathrm dx$, but this question cannot be solved through the same simple substitution of $u=\arctan x$. WolframAlpha cannot find a symbolic solution to this problem, and this Quora answer
is the only thing I can find that appears to have the exact answer of $\frac14\log^2(1+\sqrt2)$. I am not sure if there is some special trick to be used in solving this, but I have tried everything I know and nothing seems to work.
[Math] How to solve $\int_0^1\frac{\arctan(x^2)}{1+x^2}\,\mathrm dx$
definite integralsintegration
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Best Answer
The trick is to exploit a very well-hidden symmetry. The original integral equals $$ \frac{1}{2}\int_{0}^{1}\frac{\arctan(x)}{\sqrt{x}(1+x)}\,dx \tag{1}$$ which by enforcing the substitution $x\mapsto\frac{1-z}{1+z}$ becomes $$ \frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{4}-\arctan(z)}{\sqrt{1-z^2}}\,dz \tag{2}$$ so the problem boils down to evaluating $$ \int_{0}^{\pi/2}\arctan(\sin\theta)\,d\theta = \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{\pi/2}\left(\sin\theta\right)^{2n+1}\,d\theta \tag{3}$$ or the hypergeometric series $$ \sum_{n\geq 0}\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}} = \phantom{}_3 F_2\left(\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2};-1\right).\tag{4}$$ Plenty of techniques are available for such a task: in my recent works with Campbell, Cantarini and Sondow, I investigated Fourier-Legendre expansions, for instance: see 1 and 2. But the binomial transform is another perfectly viable approach. Mathematica is already able to express $\int_{0}^{\pi/2}\arctan\sin\theta\,d\theta=\int_{0}^{\pi/2}\arctan\cos\theta\,d\theta$ in terms of $\text{Li}_2$, even if my version is not able to simplify the outcome into a squared logarithm/$\text{arcsinh}$, via the usual functional equations.
The Maclaurin series of the squared (hyperbolic) arcsine is also pretty relevant here:
$$ \sum_{n\geq 0}\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}} = \int_{0}^{1}\frac{\text{arcsinh}(x)}{x\sqrt{1+x^2}}\,dx = \int_{0}^{\text{arcsinh}(1)}\frac{z}{\sinh z}\,dz\tag{5}$$ $$ \int\frac{z}{\sinh z}\,dz = z \log\sinh\frac{z}{2}+\text{Li}_2(-e^{-z})-\text{Li}_2(e^{-z})+C.\tag{6}$$
Curiously enough, this is pretty much the same principle that I exploited in this very lucky contribution of mine. And I am glad that this kind of "Eulerian" Mathematics still is a state-of-the-art, better-than-CASs Mathematics.