arctan[(1+x)/(1-x)]=π/4+∫（0到x）（arctan(1+t/1-t)）’dt为什么会有π/4?

Question

arctan[(1+x)/(1-x)]=π/4+∫（0到x）（arctan(1+t/1-t)）’dt为什么会有π/4?

Answer 1

∫（0到x）（arctan(1+t/1-t)）’dt
=F(x)-F(0)
=arctan[(1+x)/(1-x)]-π/4