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)?

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