<s>[INST] Calculate $I=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx$ How can I calculate the integral $I=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx$ by substituting $t=\frac{1-x}{1+x}$.
 [/INST] OK, the substitution you suggest works very well.  Note that
$$x = \frac{1-t}{1+t}$$
$$1+x = \frac{2}{1+t}$$
$$1+x^2 = \frac{2(1+t^2)}{(1+t)^2}$$
$$dx = -\frac{2}{(1+t)^2} dt$$
Then
$$\int_0^1 dx \frac{\log{(1+x)}}{1+x^2} = \int_0^1 dt \frac{\log{2} - \log{(1+t)}}{1+t^2}$$
With a little bit of algebra, we see immediately that
$$\int_0^1 dx \frac{\log{(1+x)}}{1+x^2} = \frac{\pi}{8} \log{2}$$
</s>