Recently a very interesting result
$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
has been proved in a more than elegant way. See Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
Here we find the value of the integral
$I=\int_{0}^1 \frac{\log(1-x)}{1+x^2} dx$
Let $x=\tan t$, then
$I=\int_{0}^{\pi/4} \log(1-\tan t) dt.$
By IVth property, we get
$I=\int_0^{\pi/4} \log (1-\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1-\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log (2 \tan x)~dx-\int_{0}^{\pi/4}\log(1+\tan x)dx=\frac{\pi}{4} \log 2+J-K.$
$J=\int_{0}^{\pi/4} \log \tan x ~dx=-C,$ see Definite integral $\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$
Let us work for $K=\int_{0}^{\pi/4} \log(1+\tan x) dx$, by IV property, again we get
$K=\int_0^{\pi/4} \log (1+\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1+\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log 2~ dx-K \implies K=\frac{\pi}{8} \log 2.$
Finally, we get $I=\frac{\pi}{8}\log 2-C,$ where $C$ is the Catalan constant.
What could be other interesting ways of finding $I$?
Best Answer
Perhaps a cleaner way is to consider the integral $$ \int_1^\infty \frac{\ln(u-1)}{1+u^2}du. $$ First, perform the substitution $u\rightarrow (u+1)/(u-1) = 1 + 2/(u-1)$ to get $$ \int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_1^\infty \frac{\ln\left(\frac{2}{u-1}\right)}{1+u^2}du = \frac{\pi}{4}\ln2 -\int_1^\infty \frac{\ln(u-1)}{1+u^2}\Longrightarrow \int_0^1\frac{\ln(u-1)}{1+u^2} = \frac{\pi}{8}\ln 2. $$ Now use the substitution $u\rightarrow x^{-1}$ to get $$ \int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_0^1\frac{\ln(x^{-1}-1)}{1+x^2}dx = \int_0^1\frac{\ln(1-x)}{1+x^2}dx -\int_0^1\frac{\ln x}{1+x^2}dx = \frac{\pi}{8}\ln 2. $$ Rearranging gives us an expression for the original integral: $$ \int_0^1\frac{\ln(1-x)}{1+x^2}dx = \frac{\pi}{8}\ln 2 + \int_0^1\frac{\ln x}{1+x^2}dx = \frac{\pi}{8}\ln 2 - C, $$ where $\int_0^1\ln(x)/(1+x^2)dx = -C$ is apparently a well-known integral.