令y=√x,代入上式得:
2∫y[（arctany）/（y(1+y^2)）]dy
=2∫[（arctany）/(1+y^2)]dy
=2∫（arctany）d(arctany)=(arctany)^2+c
=(arctan√x)^2+c
求不定积分∫[（arctan√x）/（√x(1+x)）]dx
令y=√x,代入上式得:
2∫y[（arctany）/（y(1+y^2)）]dy
=2∫[（arctany）/(1+y^2)]dy
=2∫（arctany）d(arctany)=(arctany)^2+c
=(arctan√x)^2+c