$$I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$$
Here is my attempt
$\frac{x}{1-x^2}dx=-\frac{1}{2}d\ln(1-x^2)$, integration by part, we got
$$I=-\frac{1}{2}P-Q$$
Where $P=\int_0^1 \frac{\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)}{1+x^2}dx$, and
$Q=\int_0^1 \frac{x\cdot\arctan(x)\cdot\ln(1-x^2)}{1+x^2}dx$
How to proceed then? or should I switch to other ways?
Best Answer
A (revolutionary) solution by Cornel Ioan Valean
It is one of those integrals that are very resistant to the real methods and make you wonder if you ever can do anything to get a decent (real) solution! Well, with some amount of creativity it is possible to get a very elegant solution. The integral is also included in the sequel of (Almost) Impossible Integrals, Sums, and Series (an upcoming title).
Let's observe that by symmetry we have $$I=\int _0^1\int _0^1\int _0^1\frac{x^2}{(1+x^2)(1+z^2)(1+x^2 y^2 z^2)}dzdydx$$ $$=\frac{1}{3}\int _0^1\int _0^1\int _0^1\biggr(\frac{x^2}{(1+x^2)(1+z^2)(1+x^2 y^2 z^2)}+\frac{y^2}{(1+y^2)(1+x^2)(1+x^2 y^2 z^2)}+\frac{z^2}{(1+z^2)(1+y^2)(1+x^2 y^2 z^2)}\biggr)dzdydx$$ $$\overset{\color{red}{\text{A key observation}}}{=}\frac{1}{3}\int _0^1\int _0^1\int _0^1\frac{\color{red}{(1+x^2)(1+y^2)(1+z^2)-(1+x^2y^2z^2)}}{(1+x^2)(1+y^2)(1+z^2)(1+x^2 y^2 z^2)}dzdydx$$ $$\small =\frac{1}{3}\biggr(\underbrace{\int _0^1\int _0^1\int _0^1\frac{dzdydx}{1+x^2 y^2 z^2}}_{\displaystyle \beta(3)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^3}=\pi^3/32}-\underbrace{\int _0^1\int _0^1\int _0^1\frac{dzdydx}{(1+x^2)(1+y^2)(1+z^2)}\biggr)}_{\displaystyle \pi^3/64}=\frac{\pi^3}{192}.\tag1$$
On the other hand, we get $$\small I=\int _0^1\int _0^1\int _0^1\frac{x^2}{(1+x^2)(1+z^2)(1+x^2 y^2 z^2)}dzdydx=\int _0^1\int _0^1 \frac{x^2(\pi/4-x y\arctan(x y))}{(1+x^2)(1-x^2 y^2)}dydx$$ $$\overset{xy=t}{=}\int _0^1\int _0^x \frac{x(\pi/4-t\arctan(t))}{(1+x^2)(1-t^2)}dtdx=\int _0^1\int _t^1 \frac{x(\pi/4-t\arctan(t))}{(1+x^2)(1-t^2)}dxdt$$ $$=-\frac{\pi}{8}\underbrace{\int_0^1 \frac{\log((1+t^2)/2)}{1-t^2}\textrm{d}t}_{\displaystyle -\pi^2/16}-\frac{1}{2}\color{blue}{\int_0^1 \frac{t\arctan(t) \log(2/(1+t^2))}{1-t^2}\textrm{d}t},\tag2$$ where the first integral after the last equal sign becomes trivial with the variable change $\displaystyle t\mapsto \frac{1-t}{1+t}$.
Combining $(1)$ and $(2)$, we arrive at the desired result, $$\color{blue}{\int_0^1 \frac{t\arctan(t) \log(2/(1+t^2))}{1-t^2}\textrm{d}t=\frac{\pi^3}{192}}.$$
End of story
More details, more results (and connections to other tough integrals) involving such symmetry ideas in three dimensions will be found in the sequel of (Almost) Impossible Integrals, Sums, and Series.
Here is a problem proposed by Cornel years ago in RMM that works great using such ideas as above $$\int_0^{\pi/4}\int_0^{\pi/4}\frac{(\tan^2(x)+\tan^2(y))\log(\tan(x))\log(\tan(y))\operatorname{Ti}_2(\tan(x) \tan(y))}{\tan(x)\tan(y)}\textrm{d}x\textrm{d}y$$ $$=\frac{1}{368640}\psi^5\left(\frac{1}{4}\right)-\frac{21}{32}\zeta(6)-\frac{2}{3}G^3,$$ where $\displaystyle \operatorname{Ti}_2(x)$ is the inverse tangent integral, $G$ represents the Catalan's constant, and $\psi^{(n)}$ denotes the Polygamma function.