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.
Best Answer
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} $