求arctan根号下x的不定积分,

∫arctan√xdx
=xarctan√x-∫x*1/[1+(√x)^2]*1/2*1/√xdx
=xarctan√x-1/2*∫√x/(1+x)*dx （令√x=t,则x=t^2,dx=2tdt）
=xarctan√x-1/2*∫t/(1+t^2)*2tdt
=xarctan√x-∫[1-1/(1+t^2)]dt
=xarctan√x-t+∫[1/(1+t^2)]dt
=xarctan√x-√x+arctan√x+C
=(x+1)arctan√x-√x+C