How to evaluate $\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$ How to evaluate
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$$
I completely have no idea how to find the result.Mathematic gave me the following answer part of the integral
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=-\frac{1}{4}\mathbf{G}\pi +\frac{\pi ^{2}}{16}\ln 2+\frac{21}{64}\zeta \left ( 3 \right )$$
where $\mathbf{G}$ donates the Catalan's Constant.
But it can't evaluate the other part.So I'd like to know how to evaluate the original integral or the above integral.
 A: Let $t=\arctan x$. Then
\begin{eqnarray}
&&\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\
&=&-\int_{0}^{\frac{\pi}{4}}t\ln[\cos t(\cos t+\sin t))]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\cos t^2(\cos t+\sin t)^2]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\frac{1+\cos 2t}{2}(1+\sin 2t)]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[(1+\cos 2t)(1+\sin 2t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
&=&-\frac{1}{8}\int_{0}^{\frac{\pi}{2}}t\ln[(1+\cos t)(1+\sin t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
&=&-\frac{\pi}{32}\int_{0}^{\frac{\pi}{2}}\ln[(1+\cos t)(1+\sin t)]\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
&=&-\frac{\pi}{16}\int_{0}^{\frac{\pi}{2}}\ln(1+\cos t)\mathrm{d}t+\frac{1}{64}\pi^2\ln2\\
\end{eqnarray}
Noting
\begin{eqnarray}
\int_{0}^{\frac{\pi}{2}}\ln(1+\cos t)\mathrm{d}t&=&\int_{0}^{\frac{\pi}{2}}\ln(2\cos^2\frac{t}{2})\mathrm{d}t\\
&=&\frac{\pi}{2}\ln 2+2\int_{0}^{\frac{\pi}{2}}\ln(\cos\frac{t}{2})\mathrm{d}t\\
&=&\frac{\pi}{2}\ln 2+4\int_{0}^{\frac{\pi}{4}}\ln(\cos t)\mathrm{d}t\\
&=&2G-\frac{\pi}{2}\ln 2
\end{eqnarray}
and hence
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x=\frac{1}{64} \pi  (3\pi  \ln2-8 G).$$
A: \begin{align}J&=\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\
&\overset{y=\frac{1-x}{1+x}}=\frac{\pi}{4}\int_0^1\frac{1}{1+y^2}\ln\left ( \frac{1+y^{2}}{1+y} \right )\mathrm{d}y-J\\
2J&=\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y-\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y \right )}{1+y^2}\mathrm{d}y\\
K&=\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y\\&\overset{u=\frac{1}{y}}=\int_1^{\infty}\frac{\ln\left(1+u^2\right)}{1+u^2}du-2\underbrace{\int_1^{\infty}\frac{\ln u}{1+u^2}du}_{v=\frac{1}{u}}\\
2K&=\int_0^\infty\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y+2\int_0^{1}\frac{\ln u}{1+u^2}du\\
&\overset{z=\sqrt{\frac{1-\frac{u}{\sqrt{{{u}^{2}}+1}}}{1+\frac{u}{\sqrt{{{u}^{2}}+1}}}}}=-2\int_0^1 \frac{\ln\left(\frac{4z^2}{(1+z^2)^2}\right)}{1+z^2}dz-2\text{G}\\
&=2\text{G}-\pi\ln 2+4K\\
K&=\boxed{\frac{1}{2}\pi\ln 2-\text{G}}\\
L&=\int_0^1\frac{\ln\left ( 1+y \right )}{1+y^2}\mathrm{d}y\\
&\overset{u=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(\frac{2}{1+u}\right)}{1+u^2}du\\
2L&=\boxed{\frac{1}{4}\pi\ln 2}\\
\end{align}
Therefore,
\begin{align}\boxed{\displaystyle J=\dfrac{3}{64}\pi^2\ln 2 -\dfrac{1}{8}\pi\text{G} }\end{align}
