I solved this using the following technique. But I prefer simple techniques for short questions!
Let $x=\frac{1-t}{1+t},$ we have $\frac{dx}{dt}=-\frac{2}{(1+t)^2}.$
Thus $\int_0^1 \frac{\ln (x+1)}{x^2+1}dx=\int_1^0 \frac{\ln \frac{2}{1+t}}{\left(\frac{1-t}{1+t}\right)^2+1}\cdot \left\{-\frac{2}{(1+t)^2}\right\} dt$
$=2\int_0^1 \frac{\ln \frac{2}{1+t}}{(1+t)^2+(1-t)^2}\ dt$
$=\int_0^1 \frac{\ln 2-\ln (1+t)}{t^2+1}\ dt$
$=\int_0^1 \frac{\ln 2}{x^2+1}\ dx-\int_0^1 \frac{\ln (x+1)}{x^2+1}\ dx\ (\because \ t \ is \ an \ arbitrary \ real \ number)$
$\Longleftrightarrow 2\int_0^1 \frac{\ln (x+1)}{x^2+1}\ dx=\ln 2\cdot \int_0^1 \frac{1}{x^2+1}\ dx=\frac{\pi}{4}\ln 2$
$\therefore \int_0^1 \frac{\ln (x+1)}{x^2+1}\ dx=\frac{\pi}{8}\ln 2.$
