Interestingly, I seem to have an integral I have posted before, but I want to take a different approach to it.
$\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \,dx$
The beta function states,
$B(x,y) = \int_{0}^{1} {t}^{x-1}({1-t}^{y-1}) \,dx$
So, I was just thinking [B]if[/B] there a possible way to compute this integral using the beta function also knowing
$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
Best Answer
I am thinking about a creative use of the Euler Beta function.
For now, just a symmetry trick related to the formula $\tan(\pi/4-u)=\frac{1-\tan u}{1+\tan u}$:
$$\begin{eqnarray*}I&=&\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,dx = \int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta\\&=&\int_{0}^{\pi/4}\log\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)\,d\theta=\frac{\pi}{4}\log 2-I\end{eqnarray*}$$ from which: $$ I = \color{red}{\frac{\pi}{8}\log 2}.$$