∫ arctan(1+√x)dx
换元t=arctan(1+√x),(tant -1)^2=x
=∫ t d(tant-1)^2
=t(tant-1)^2 - ∫ (tant-1)^2 dt
=t(tant-1)^2 - ∫ (sint-cost)^2/cos^2t dt
=t(tant-1)^2 - ∫ (1-2sintcost)/cos^2t dt
=t(tant-1)^2 - ∫ 1/cos^2t dt + 2∫ sint/cost dt
=t(tant-1)^2 - tant - 2∫ 1/cost d(cost)
=t(tant-1)^2 - tant - 2ln|cost| + C
=x*arctan(1+√x)-(1+√x)-2ln|cos(arctan(1+√x))| + C
有不懂欢迎追问