How do I integrate $\int_0^1\arctan(x)\log(\frac{1-x}{1+x})\mathrm{d}x$? I've recently come across an interesting integral, which is of the form:
$$\int_0^1\arctan(x)\log\left(\frac{1-x}{1+x}\right)\mathrm{d}x$$
To start, I expanded the arctangent into its series expansion, then utilized the Weierstraß substitution in order to remove the fractional term from the logarithm:
$$t = \frac{1-x}{1+x}$$
Finally, I'm left with this integral:
$$2 \sum_{k \geq 0} \frac{(-1)^k}{2k+1} \int_0^1 \frac{(1-t)^{2k+1}}{(1+t)^{2k+3}} \log(t)\mathrm{d}t$$
Which looks an awful lot like the beta function, namely:
$$B(x, y) = (1-a)^y \int_0^1 \frac{(1-t)^{x-1} t^{y-1}}{(1-at)^{x+y}} \mathrm{d}t, \quad a \leq 1$$
For the following values, the integrals are nearly identical:
$$a=-1,$$
$$x=2k+2,$$
$$y=1$$
However, this is the bit where I fail to make progress.
I see that the integrals are clearly just off by that logarithm, but I cannot find a relation between them in order to progress with this integral.
I've tried differentiating with respect to the parameter $y$ in order to bring in that logarithm, but that obviously doesn't do much - as the parameter also lies in the denominator and causes unwanted trouble.
I've also tried constructing integrals which are similar to this one, but only have the parameter $y$ in the numerator; however, I haven't been able to make much progress doing that either. These integrals end up looking nothing like the beta function.
 A: Integrate by parts with $dx=d(x-1)$
\begin{align}
&\int_0^1\tan^{-1}x\ln\frac{1-x}{1+x}\ {dx}\\
\overset{ibp}= & \int_0^1 \frac{(1-x)\ln \frac{1-x}{1+x}}{1+x^2}\ \overset{\frac{1-x}{1+x}\to x}{dx}
-2\int_0^1 \frac{\tan^{-1}x}{1+x}dx\\
=& \ \int_0^1 \frac{\ln x}{1+x^2}dx -\frac34 \int_0^1\frac{\ln x}{1+x}dx
 -2\int_0^1 \frac{\tan^{-1}x}{1+x}dx\\
=& -G-\frac34 \left(-\frac{\pi^2}{12}\right)-2\cdot \frac\pi8\ln2
=-G +\frac{\pi^2}{16} -\frac\pi4\ln2
\end{align}
where $\int_0^1 \frac{\ln x}{1+x^2}dx=-G$ and $\int_0^1\frac{\ln x}{1+x}dx= -\frac{\pi^2}{12} $
A: Noting
$$ d\bigg[(x-1)\ln\bigg(\frac{1-x}{1+x}\bigg)-2\ln(1+x)\bigg]=\ln\bigg(\frac{1-x}{1+x}\bigg) $$
one has, by IBP,
\begin{eqnarray}
&&\int_0^1\tan^{-1}x\ln\bigg(\frac{1-x}{1+x}\bigg)\ {dx}\\
&=&-\int_0^1\bigg[(x-1)\ln\bigg(\frac{1-x}{1+x}\bigg)-2\ln(1+x)\bigg]\frac{1}{1+x^2}dx\\
&\overset{x\to\frac{1-x}{1+x}}{=}&\int_0^1\bigg(\frac{\ln x}{1+x^2}+\frac{x\ln x}{1+x^2}-\frac{\ln x}{1+x}+\frac{2\ln(\frac2{1+x})}{1+x^2}\bigg)dx\\
&=&-G-\frac{\pi^2}{48}+\frac{\pi^2}{12}+\frac14\pi\ln2\\
&=&-G+\frac{\pi^2}{16}+\frac14\pi\ln2.
\end{eqnarray}
Update: Let
\begin{eqnarray}
I&=&\int_0^1\frac{\ln(1+x)}{1+x^2}\\
&=&\int_0^1\int_0^1\frac{x}{(1+tx)(1+x^2)}dtdx\\
&=&\int_0^1\int_0^1\frac{x}{(1+tx)(1+x^2)}dxdt\\
&=&\int_0^1\bigg(\frac\pi4\frac{t}{1+t^2}+\frac{2\ln2}{1+x^2}-\frac{1+t}{1+t^2}\bigg)dt\\
&=&\frac14\pi\ln2-I
\end{eqnarray}
and hence
$$ I=\frac18\pi\ln2. $$
\begin{eqnarray}
&&\int_0^1\frac{2\ln(\frac2{1+x})}{1+x^2}\\
&=&\frac{\pi}2\ln2-2I=\frac{\pi}4\ln2.
\end{eqnarray}
