Prove $\int_0^{\infty} \frac{\arctan{(x)}}{x} \ln{\left(\frac{1+x^2}{{(1-x)}^2}\right)} \; \mathrm{d}x = \frac{3\pi^3}{16}$ Prove that $$\int_0^{\infty} \frac{\arctan{(x)}}{x} \ln{\left(\frac{1+x^2}{{(1-x)}^2}\right)} \; \mathrm{d}x = \frac{3\pi^3}{16}$$
This is not a duplicate of this post, the bounds are different and the integral evaluates to a slightly different value.  I tried looking at the solution from the linked post but I'm not familiar with harmonic numbers or complex analysis and the real solution is long.  I tried IBP but got no where.  Any advice for this monster integral (real analysis only please)?
 A: Changing the bounds makes the integral way simpler, because after letting $x\to \frac{1}{x}$ we can get rid of that $\arctan x$.
$$I=\int_0^{\infty} \frac{\arctan x}{x} \ln\left(\frac{1+x^2}{{(1-x)}^2}\right)dx\overset{x\to \frac{1}{x}}=\int_0^\infty \frac{\arctan \left(\frac{1}{x}\right)}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx$$
$$\Rightarrow 2I=\frac{\pi}{2} \int_0^\infty \frac{1}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{x = \tan \frac{t}{2}}=-\frac{\pi}{2}\int_0^\pi\frac{\ln(1-\sin t)}{\sin t}dt$$
Also from here we know that:
$$I(a)=\int_{0}^{\pi} \frac{\ln(1+\sin a\sin x)}{\sin x}dx=a(\pi -a)$$
$$\Rightarrow I=-\frac12 \frac{\pi}{2}I\left(\frac{3\pi}{2}\right)=-\frac12 \frac{\pi}{2}\left(-\frac{3\pi^2}{4}\right)=\frac{3\pi^3}{16}$$

Another way to deal with the last integral (credits to this answer), is to consider:
$$\mathcal J(a)=\int_0^\frac{\pi}{2}\arctan\left(\frac{\sin x -\tan\frac{a}{2}}{\cos x}\right)dx$$
And differentiate w.r.t. a, obtaining:
$$\mathcal J'(a)=-\frac12\int_0^\frac{\pi}{2}\frac{\cos x}{1-\sin a\sin x}dx=\frac12 \frac{\ln(1-\sin a)}{\sin a}$$
$$\mathcal J(\pi)-\mathcal J(0)=-\frac{\pi^2}{4}-\frac{\pi^2}{8}=\frac12\int_0^\pi\frac{\ln(1-\sin a)}{\sin a}da$$
$$\Rightarrow \int_0^\pi \frac{\ln(1-\sin a)}{\sin a}da=-\frac{3\pi^2}{4}$$
A: Enforce the substitution $x\mapsto 1/x$ and use $\arctan(1/x)=\pi/2-\arctan(x)$ to find that
$$\begin{align}
\color{blue}{\int_1^\infty \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx}&=\int_0^1 \left(\frac{\pi/2-\arctan(x)}{x}\right)\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx\\\\
&=\frac\pi2 \int_0^1 \frac{\log(1+x^2)}{x}\,dx-\pi\int_0^1\frac{\log(1-x)}{x}\,dx\\\\
&-\color{blue}{\int_0^1\frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx}\\\\
\color{blue}{\int_0^\infty \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx}&=\frac\pi2 \int_0^1 \frac{\log(1+x^2)}{x}\,dx-\pi\int_0^1\frac{\log(1-x)}{x}\,dx
\end{align}$$

Now, expanding $\log(1+x)$ in its Taylors series and integrating term by term reveals that
$$\begin{align}
\int_0^1 \frac{\log(1+x^2)}{x}\,dx&=\frac12\int_0^1 \frac{\log(1+x)}{x}\,dx\\\\
&=\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\\\\
&=\frac{\pi^2}{24}
\end{align}$$
and similarly that
$$\int_0^1\frac{\log(1-x)}{x}\,dx=-\frac{\pi^2}{6}$$

Putting it together, we find the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)\,dx=\frac{3\pi^3}{16}}$$
